L 


SOLID 

GEOMETRY 

STONE- MiLLIS 


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BENJ.H,  SANBORN  f 


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IN  MEMORIAM 
FLORIAN  CAJORl 


SOLID  GEOMETRY 


BY 


JOHN   C.    STONE,   A.M. 

)) 

HEAD      OF     THE     DEPARTMENT     OF      MATHEMATICS,      STATE      NORMAL      SCHOOL 

MONTCLAIR,  NEW  .JERSEY,   CO-ACTHOR  OF  THE  80UTHW0RTH-ST0NE 

ARITHMETICS,   STONE-MILLIS  ARITHMETICS,  SECONDARY 

ARITHMETIC,    ALGEBRAS    AND    GEOMETRIES 

AND 


JAMES   F.    MILLIS,   A.M. 


HEAD    OF   THE    DEPARTMENT    OF    MATHEMATICS,    FRANCIS    W.    PARKER    SCHOOI 

CHICAGO,    CO-AUTHOR    OF    THE    STONE-MILLIS    ARITHMETICS 

SECONDARY    ARITHMETIC,    ALGEBRAS 

▲ND   G£OM£TRI£S 


ov  TToXX'  dWa  iroXv 


BENJ.    H.    SANBORN   &   CO. 

CHICAGO  NEW   YORK  BOSTON 

1925 


STONE-MILLIS  MATHEMATICS   SERIES 


LEMENTAKY   ARITHMETICS: 

Primary 

Primary 

Intermediate 

Complete 

Advanced 

Secondary  Arithmetic 

Elementary  Algebra,  First  Course 

Elementary  Algebra,  Second  Course 

Elementary  Algebra,  Complete  Course 

Higher  Algebra 

Plane  Geometry 

Solid  Geometry 

Plane  and  Solid  Geometry 


Benj.  H.  Sanborn  &  Co.,  Publishers 


OOPTBIGHT,   1916 
BY 

BENJ.   H.  SANBORN  4  00. 


PREFACE 

The  Stone-Millis  Geometry  —  Plane,  Solid,  and  Plane 
AND  Solid  —  published  in  1910,  was  a  pioneer  in  its  field,  being 
the  first  of  the  group  of  American  textbooks  on  geometry  which 
in  recent  years  have  attempted  in  various  ways  to  make  the 
teaching  of  geometry  conform  to  modern  thought  in  education. 
It  marked  a  wide  departure  from  the  traditional  Greek  geometry 
after  which  textbooks  for  secondary  schools  had  for  generations 
been  patterned.  This  text  has  met  with  remarkable  success. 
The  educational  ideals  which  it  embodied  are  now  recognized  as 
national,  and  are  summarized  in  the  Report  of  the  National 
Committee  of  Fifteen  on  the  teaching  of  geometry. 

The  present  geometry,  by  the  same  authors,  has  been  prepared 
in  the  attempt  to  produce  a  text  which  shall  preserve  the  dis- 
tinctive features  of  the  older  text,  but  which,  if  possible,  shall 
be  more  simple,  practical,  and  teachable. 

The  following  are  some  of  the  features  which  distinguish  this 
text: 

1.  Simplicity.  —  1.  The  subject  is  graded  so  that  the  easier 
topics  come  first  and  so  that  the  student  is  introduced  to  only 
one  new  difficulty  at  a  time.  The  grading  of  geometry  is  made 
possible  in  this  text  by  abandoning  the  Greek  division  of  geome- 
try into  books  and  re-grouping  the  material  in  chapters. 

2.  Some  of  the  theorems  on  fundamental  properties  of  figures 
are  treated  informally  at  the  beginnings  of  many  topics. 

3.  The  subject  is  also  abbreviated  and  simplified  by  the  omis- 
sion of  certain  useless  traditional  theorems  and  the  reduction  to 
a  reasonable  minimum  of  the  number  of  theorems,  constructions, 
and  corollaries  requiring  formal  treatment. 

iii 


Mr?05027 


IV  PREFACE 

4.  Tlie  use  of  the  theory  of  limits  in  the  proofs  of  incommen- 
surable cases  of  theorems  has  been  eliminated. 

II.  Practicality.  —  1.  Geometry  is  humanized  by  using  as 
exercises  a  large  number  of  practical  problems.  This  phase  of 
geometry,  which  was  first  introduced  into  American  schools  by 
the  Stone-Millis  text,  is  now  universally  recognized  as  an  integral 
part  of  the  subject.  The  Stone-Millis  Geometry  contains  a 
very  large  number  and  a  very  great  variety  of  simple  and  genu- 
ine practical  problems.  They  are  selected  from  many  fields  of 
human  activity,  such  as  home  life,  art,  architecture,  astronomy, 
engineering,  designing,  navigation,  science,  the  construction  and 
use  of  implements  and  machinery,  etc. 

2.  Directions  are  given  for  the  construction  of  many  home- 
made instruments  and  their  use  in  out-of-door  exercises. 

3.  Geometry  is  correlated  with  trigonometry  by  the  introduc- 
tion of  simple  work  with  trigonometric  ratios,  in  the  chapter  on 
similar  polygons.  Application  is  made  to  practical  problems  in 
the  solution  of  triangles. 

III.  Teachableness.  —  1.  A  concrete  approach  to  formal 
geometry  is  provided.  This  develops  a  body  of  experience  and 
imagery  as  a  basis  of  formal  geometry,  and  the  latter  is  not 
introduced  until  need  for  it  is  felt.  In  the  approach  to  demon- 
strative geometry,  familiarity  with  important  geometric  figures 
is  secured  through  their  accurate  construction  with  drawing  in- 
struments. This  development  of  clear  imagery  through  accurate 
drawing  of  the  figures  involved  in  the  formal  theorems,  etc.,  is 
continued  throughout  plane  geometry. 

2.  Use  is  made  of  the  suggestive  method  in  the  treatment  of 
theorems.  While  complete  model  proofs  of  a  large  number  of 
theorems  are  given  —  and  whenever  a  proof  is  given  it  is  given 
in  complete  form,  with  numbered  steps  —  the  proofs  of  many 
theorems  are  left,  with  suggestions,  to  the  student.  The  sug- 
gestions in  a  large  part  of  the  theorems  are  given  in  the  form 
of  analyses.  It  is  believed  that  suggestions  of  this  nature  are 
superior  to  those  of  the  traditional  kind,  which  are  mere  outlines 


PREFACE  V 

of  tlie  proofs,  because  they  give  the  student  training  in  exactly 
the  kind  of  thinking  which  he  must  do  when  attacking  a  proof 
unaided,  and  thus  they  teach  method  of  attack  and  develop  power 
of  originality. 

3.  Exercises  which  demand  technical  knowledge  have  been 
eliminated.  Many  new  exercises  have  been  introduced.  The 
exercises  are  grouped  in  every  case  immediately  after  the  theo- 
rem, construction,  or  corollary  to  which  they  relate.  Many  mis- 
cellaneous review  exercises  are  given. 

4.  Special  care  has  been  given  to  the  illustrations  in  this  text. 
When  construction  lines  are  required  in  drawing  a  figure,  they 
show  in  the  book.  Throughout  the  Solid  Geometry  shaded 
drawings  of  models  are  placed  by  the  side  of  the  more  compli- 
cated geometric  figures,  to  aid  the  student  in  visualizing  the  third 
dimension  in  the  figures  while  looking  at  the  flat  drawings  of  them. 
The  consistent  plan  of  representing  hidden  parts  of  figures  by 
thinner  lines  than  the  others  is  carried  out,  dotted  lines  being 
employed  exclusively  for  auxiliary  lines  as  in  plane  geometry. 

Grateful  acknowledgment  of  the  authors  is  due  to  all  those 
who  by  timely  suggestions  have  aided  in  the  preparation  of  this 
text;  especially  to  Professor  H.  E.  Cobb  and  Professor  A.  W. 
Cavanaugh  of  Lewis  Institute,  Chicago ;  to  Miss  Alice  M.  Lord 
of  the  High  School,  Portland,  Maine ;  and  to  Professor  Guido  H. 
Stempel  of  Indiana  University.  Special  acknowledgment  is  due 
to  Mr.  Charles  McCauley  of  Chicago,  who  has  made  the  excellent 

illustrations. 

JOHN  C.    STONE, 
JAMES  F.    MILLIS. 
Januabt,  1916. 


CONTENTS 

CHAPTER  PAGE 

Xil.    Lines  and  Planes  in  Space      ...                .        .  277 

XUI.    Prisms  and  Cylinders       .        .        .        .        .        .        .  321 

XIV.     Pyramids  and  Cones  ........  355 

XV.     Regular    Polyedrons.      Similar  Polyedrons.      Pris- 

MATOIDS    . .  390 

XVI.    The  Sphere 404 


I 


▼U 


SYMBOLS 


=,  is  equal  to;  equals. 

=,  is  identically  equal  to. 

~,  is  similar  to. 

^,  is  congruent  to. 

>,  is  greater  than. 

<,  is  less  than. 

=  ,  approaches  as  a  limit. 

II  ,  is  parallel  to ;  parallel. 

The  plural  of  any  symbol  representing  a  noun  is  obtained  by  affixing 
the  letter  s.    Thus,  A  represents  angles. 


±,  is  perpendicular  to;    perpen- 
dicular. 
.*.,  therefore. 
•••,  and  so  on. 
Z.,  angle. 
A,  triangle. 
O,  parallelogram. 
O,  circle. 


ABBREVIATIONS 


Ax.y  axiom. 
Alt.,  alternate. 
Comp.,  complementary. 
Cor.,  corollary. 
Carres.,  corresponding. 
Def.,  definition. 
Ex.f  exercise. 


Ext.,  exterior. 
Hyp.,  hypothesis. 
Int.,  interior. 
Rect.,  rectangle. 
Rt.,  right. 
St.,  straight. 
Supp.,  supplementary. 


SOLID   GEOMETRY 


CHAPTER   XII 


LINES   AND   PLANES   IN   SPACE 


288.  Solid  geometry.  —  It  has  been  seen  that  plane  geometry 
deals  with  figures  which  lie  in  a  flat  or  plane  surface.  Solid 
geometry  treats  of  figures  consisting  of  geometric  solids,  sur- 
faces, lines,  points,  or  combinations  of  them,  which  are  not 
confined  to  a  plane. 

Solid  geometry,  like  plane  geometry,  has  been  developed 
by  man  in  the  endeavor  to  meet  his  practical  needs,  and 
dates  back  to  an- 
cient times.  Ahmes, 
an  Egyptian,  in 
his  book  entitled 
"  Directions  for  Ob- 
taining the  Knowl- 
edge of  All  Dark 
Things,"  written 
about  1700  B.C., 
gave  rules  for  find- 
ing the  contents  of 
wells  and  granaries.  The  principal  geometric  forms  — 
prismatic,  cylindrical,  pyramidal,  conical,  and  spherical  — 
appear  in  nature  in  endless  combinations.  They  are 
employed  by  man  in  countless  ways  —  in  the  construction 
of   buildings,   monuments,  temples,  embankments;    in   the 

2:7 


Lincoln  Memorial,  Washington,  D.  C. 


278 


SOLID   GEOMETRY 


making  of  pottery,  jewelry,  furniture ;  in  the  great  fields 
of  engineering,  mechanics,  architecture,  art,  astronomy, 
etc. 

289.  Geometric  drawings  and  models.  —  Although  the  figures 
of  solid  geometry  are  not  in  one  plane,  they  must  be  repre- 
sented by  drawings  all  parts  of  which  do  lie  in  one  plane. 
It  is  necessary,  therefore,  for  the  student  of  solid  geometry 
to  learn  to  sense  thickness,  or  the  third  dimension,  of  a  figure 
while  looking  at  a  flat  drawing  of  it.  For  .assistance  in 
overcoming  this  difficulty  of  the  third  dimension,  those 
parts  of  a  geometric  figure  which  would  be  invisible  if  it 
were  a  physical  object  are  represented  in  a  drawing  by 
thinner  lines  than  the  other  lines  of  the  figure.  This  is 
illustrated  in  the  drawing  of  a  prism  below.  When  auxiliary 
lines  occur,  they  are  represented  by  dotted  lines  as  in  plane 
geometry. 


<_> 


^j—b" 


Peism 


Also,  for  assistance  in  sensing  the  third  dimension  in  a 
drawing,  sometimes  a  picture  of  a  model  of  the  figure  is 
given  in  the  text,  adjoining  the  geometric  figure,  as  illus- 
trated in  the  case  of  the  prism  above. 

The  use  of  wooden,  wire,  or  cardboard  models  would  be 
of  great  aid  to  the  beginning  student.  Models  may  be  con- 
structed of  cardboard  as  illustrated  in  the  following  exer- 
cises. 


LINES  AND   PLANES  IN   SPACE 


279 


EXERCISES 

1.  On  a  piece  of  cardboard,  draw  a  figure 
similar  to  the  adjoining  figure,  making  the 
side  of  each  square  3  in.  Cut  out  the  pat- 
tern, and  by  folding  along  the  dotted  lines 
and  pasting,  make  a  model  of  a  cube. 

How  many  squares  form  the  surface  of 
this  model  ?  How  many  edges  has  it  V  How 
many  corners  has  it  ? 

2.  On  a  piece  of  cardboard,  draw  a  figure 
similar  to  the  adjoining  figure,  making  the 
side  of  each  hexagon  2  in.  Cut  out  the 
pattern,  and  by  folding  along  the  dotted 
lines  and  pasting,  make  a  model  of  a  hex- 
agonal prism. 

Models  of  other  kinds  of  prisms  may  be 
made  by  using  other  kinds  of  regular  poly- 
gons instead  of  the  hexagons. 

3.  Following  the  method  of  Exercises  1 
and  2,  by  first  making  a  pattern  similar  to 
the  pattern  here  shown,  with  the  side  of 
each  triangle  4  in.,  construct  a  model  of  a 
triangular  pyramid. 


ff=h 


ri — ' — I    i  ^r- 

I  t       I 

I  III 

I  III 

j        r  II 

N        ! I ? 


290.  Representation  of  a  plane  surface  —  A  plane  surface  is 
unlimited   in    extent.     Hence    it    is    impossible     to    show 
an   entire   plane   to   the   eye   by  a 
drawing.      In     geometric     figures, 
a  plane  is  represented  by  some  kind 
of  polygon,  which  incloses  a  portion     ^^ 
of  the  plane.     Trapezoids  and  parallelograms  are  used  ex- 
tensively for  representing  planes. 

This  figure  represents  a  horizontal  plane,  seen  obliquely. 
The  plane  is  named  MN. 


280  SOLID   GEOMETRY 

291.  Fundamental  property  of  a  plane  surface.  — A  straight- 
edge, held  so  that  it  touches  a  flat  or  plane  surface  at  two 
points,  touches  the  surface  all  along  the  straightedge. 
Hence  it  is  inferred  that : 

(1)  If  two  points  of  a  straight  line  lie  in  a  plane^  the  entire 
line  lies  in  the  plane. 

From  the  fundamental  property  above,  let  the  student 
prove  the  following : 

(2)  A  straight  line  can  intersect  a  plane  in  only  one  point. 

292.  Revolution  of  a  plane.  —  It  ib  evident  that : 

By  revolving  a  plane  about  any  straight  line  in  it  as  an  axis, 
it  may  be  made  to  contain  any  point  of  space  that  is  not  on  the 
line,  in  one  and  only  one  position. 

Thus,  ii  AB  is  a  straight  line  in 
plane  MN  and  C  is  any  point  of 
space  that  is  not  on  AB,  by  revolving 
MN  about  AB  as  an  axis,  it  will  arrive 
at  a  position  PQ  in  which  it  contains 
point  C.  If  the  plane  is  revolved  far- 
ther, it  will  no  longer  contain  point  C. 

293.  Corollary  l.  —  A  straight  line  lies  in  an  unlimited 
number  of  planes. 

For,  a  straight  line  AB  lies  in  at  least  one  -plane,  by 
definition  of  a  straight  line.  (See 
§  6.*)  This  plane  may  be  revolved 
about  AB,  by  §  292.  And  each 
successive  position  of  the  revolving 
plane  represents  a  different  plane 
containing  AB. 


r 

^C 

I  A 

^,^-^ 

yN 

M\^ 

^ 

B 

0 

*  See  the  References  to  Plane  Oeometry  at  the  end  of  the  book. 


LINES  AND   PLANES   IN   SPACE  281 

294.  Corollary  2.  —  An  unlimited  number  of  straight  lines 
may  he  drawn  perpendicular  to  a  given  straight  line  in  space  at 
a  given  point  of  the  line. 

For,  at  any  point  of  the  line  AB  in  §  298,  a  perpendicular 
to  AB  may  be  drawn  in  each  of  the  planes  containing  AB. 

295.  Detennming  a  plane.  —  A  plane  is  determined  by  cer- 
tain points,  lines,  or  combinations  of  them,  if  that  plane  and 
no  other  plane  contains  all  of  those  points,  etc. 

296.  Theorem.  —  A  straight  line  and  a  point  not  on  the  line 
determine  a  plane. 


Hypothesis.     AB  is  any  straight  line  and  O  any  point  of 
space  not  on  AB, 

Conclusion.     Line  AB  and  point  C  determine  a  plane. 
Proof.     1.    (7  is  a  point  not  on  straight  line  AB.         Hyp. 

2.  AB  lies  in  a  plane.  §  6 

3.  By  revolving  this  plane  about  AB  as  an  axis,  it  may  be 
made  to  contain  point  O  in  one  and  only  one  position.    §  292 

4.  Hence  line  AB  and  point  C  determine  a  plane.      §  295 
Without  the  book,  draw  a  figure  and  write  out  this  proof. 

297.    Corollary  l.  —  Three  points  not  in  the  same  straight  line 
determine  a  plane.  ^ 

For,  if  J.,  B,  and  Q  are  the  three  /  '^^ 

points,   A    and    B    determine    a       /   ^)». -^^ 

straight  line.     One  and  only  one       

plane  can  contain  line  AB  and  point  C,  by  §  296.     Hence, 
one  and  only  one  plane  can  contain  J.,  B,  and  C.     Why  ? 


282  SOLID   GEOMETRY 

298.  Corollary  2.  —  Two  intersecting  straight  lines  determine 
a  plane. 

For,  if  the  straight  lines  AB  and  QB  intersect  at  0,  let  P 
be  any  other  point  on  OB,     Then 
one  and  only  one  plane  can  contain         /    ^ 

AB  and  point  P,  by  §  296.     This      /    C^...^^^^^ ^ 

plane  contains  the  whole  line  (7i>, 

by  §  291.     Why  can  only  this  one  plane  contain  the  lines  AB 

and  QB'l 

299.  Corollary  3.  —  Two  'parallel  straight  lines  determine  a 
plane. 


For,  if  AB  and  OB  are  parallel,  /     ^     ^p 

they  lie  in  a  plane,  by  definition 
of  parallel  lines.     Let  P  be   any 
point  on  AB.    Then  only  one  plane 
can  contain  point  P  and  OB.     Why  ?     Why,  then,  can  only 
one  plane  contain  AB  and  OB  ? 

EXERCISES 

1.   What  tool  does  a  carpenter  use  for  reducing  the  rough  surface  of 
a  board  to  a  smooth  or  plane  surface  ?     Explain 
how  the  principle  in  §  291  is  involved  in  the  use 
of  this  tool. 


2.  A  plasterer,  after  putting  plaster  on  a 
wall,  smooths   it   down   to  a  plane  surface  by 

rubbing  a  board  or  straightedge  over  it  in  many  directions.  Builders  of 
concrete  sidewalks  do  the  same  thing.  Why  does  this  produce  a  plane 
surface  ? 

3.  In  measuring  grain  with  a  half-bushel  measure,  the  grain  is  first 
heaped,  then  it  is  raked  off  even  with  the  top  of  the  measure,  with  a 
board.  Why  is  the  measure  then  level  full  ?  What  kind  of  surface  is 
thus  formed  by  the  grain  ? 


LINES  AND  PLANES  IN   SPACE 


283 


4.  Illustrate  the  principle  in  §  292  by 
using  the  point  of  a  pencil  to  denote  any  point 
of  space  and  a  page  of  an  open  book  to  denote 
the  revolving  plane.  Think  of  another  way 
in  which  this  principle  may  be  illustrated. 

5.  Place  three  tacks  on  a  table  so  that  they  are  not  in  a  straight  line 
and  have  their  points  up.  Support  a  book  or  a  pane  of  glass  on  the 
points  of  the  tacks.  How  does  this  illustrate 
§297? 

6.  Place  a  ruler  on  a  table,  with  its  edge  up, 
and  place  a  tack  on  the  table,  with  its  point  up. 
Support  a  book  or  a  pane  of  glass  on  the  edge  of 
the  ruler  and  the  point  of  the  tack.  How  does 
this  illustrate  §  296  ? 

7.  Place  two  rulers  in  parallel  positions  on  a 
table,  with  their  edges  up.     Support  a  book  or 

pane  of  glass  on  the  edges  of  the  rulers.     How  does  this  illustrate  §  299  ? 

8.  Cameras,  telescopes,  etc.,  are  mounted  on  tripods.     WUl  an  object 
mounted  on  three  legs  always  rest  firmly  on  the  floor?    Why? 

9.  Do  four  given  points  lie  in  one  plane  ? 

10.  Why  do  not  all  chairs,  tables,  etc.,  with  four  legs  rest  firmly  on 
the  floor  ? 

11.  Find  four  points  in  the  room  through  which  a  plane  may  be 
passed.     Find  four  points  through  which  a  plane  cannot  be  passed. 

12.  Hold  two  pencils  in  such  positions  as  to  show  that  a  plane  cannot 
be  passed  through  any  two  straight  lines  taken  at  random. 

13.  Show  that  if  a  piece  of  cord  is  fastened  at  both  ends,  and,  by 
grasping  it  at  any  third  point,  the  two  parts  are  stretched  straight,  the 
stretched  cord  will  lie  entirely  in  one  plane. 

14.  Show  that  any  transversal  of  two  parallel  lines  must  lie  in  the 
plane  of  those  lines. 

15.  Show  that  a  figure  made  of  three  straight  lines,  each  intersecting 
the  other  two,  but  not  in  a  common  point,  must  lie  in  one  plane. 


300.    Intersection  of  planes.  —  It  is  assumed  that  : 

If  two  planes  intersect,  they  have  more  than  one  point  in 


common. 


284 

301.    Theorem. 

line. 


SOLID   GEOMETRY 
The  intersection  of  two  planes  is  a  straight 


Hypothesis.     Planes  iHfiV  and  PQ  intersect. 
Conclusion.     The  intersection  of  planes  MN  and  P§  is  a 
straight  line. 

Proof.     1.    ilifiV  and  P§  intersect.  Hyp. 

2.  .  •.  iHfiV and  PQ  have  at  least  two  common  points.  Let 
them  be  A  and  B,  §  300 

3.  Then  all  points  of  the  straight  line  joining  A  and  B  lie 
in  both  planes.  §  291 

4.  No  point  without  line  AB  can  lie  in  both  planes,  for  the 
two  planes  would  then  coincide.  §  296 

5.  Hence,  since  all  common  points  of  the  planes,  and  no 
other  points,  must  lie  in  the  intersection,  line  AB  is  the  in- 
tersection of  the  planes.  That  is,  the  intersection  of  the 
planes  is  a  straight  line. 

Write  out  the  proof  without  using  the  book. 


EXERCISES 

1.  What  is  the  meaning  of  "Two  intersecting  planes  determine  a 
straight  line  "  ? 

2.  What  is  the  locus  of  all  points  common  to  two  planes? 

3.  Show  that  if  a  sheet  of  cardboard,  or  other  kind  of  stiff  sheet  ma- 
terial, is  folded  and  creased,  the  crease  must  be  straight. 

4.  When  one  saws  a  board  in  two,  why  is  the  edge  of  the  cut  made 
by  the  saw  a  straight  line? 


LINES   AND   PLANES   IN   SPACE 


285 


5.  In  slicing  a  loaf  of  bread,  if  a  surface  of 
the  loaf  is  flat,  why  is  the  edge  of  the  slice 
straight?  Think  of  other  applications  in  the 
home  of  the  principle  in  §  301. 

6.  Carpenters,  in    shingling   a  roof,    often 
mark  off  straight  chalk  lines  on  the  roof  to 
assist  in  placing  the  shingles  in  straight  rows. 
A  cord,  whitened  with  chalk,  is  held  firmly  at 
the   two    ends,   and    stretched    straight.     Then 
grasping  the  cord  at  some  third  point,  they  pull 
it  away  a  little  from  the  roof  and  let  it  go.     It 
springs   back   and   strikes   the  roof,   leaving   a 
white  mark,  which  is   a  straight  line.     Prove 
that  this  is  an  application  of  §  301. 

7.  When  three  planes,   not  containing   the 

same  straight  line,  intersect  in  pairs,  if  the  three  lines  of  intersection  arn 
not  parallel,  they  meet  at  one  point. 

Suggestions.  —  Let  the  planes  in- 
tersect in  AB,  CD,  and  EF.  Prove 
that  AB  and  CD  intersect  at  some 
point  0.  Prove  that  0  is  on  EF  by 
showing  that  it  is  in  plane  EC  and 
also  in  plane  A  E. 

8.  Illustrate  the  principle  in  Exercise  7  by  the  walls  and  fseilings  of 


-jj^^^n 

1      1  11 

II   II   II    11    II    II    1 

ll  \  1  1   11    11   11 ' 

'yvw  1  IT 

II        II  II  .1 

the  schoolroom.     Give  other  illustrations. 


302.  Line  and  plane  perpendicular.  —  A  straight  line  which 
intersects  a  plane  is  perpendicular  to  the  plane  if  it  is  perpen- 
dicular to  every  straight  line  in  the  plane  drawn  through 
the  point  of  intersection. 

The  plane  also  is  said  to  be  perpendicular  to  the  line. 

If  a  line  is  perpendicular  to  a  plane,  the  point  of  inter- 
section of  the  line  and  plane  is  called  the  foot  of  the  per- 
pendicular. 

If  a  straight  line  intersects  a  plane,  but  is  not  perpendicu- 
lar to  it,  the  line  is  said  to  be  oblique  to  the  plane. 


286 


SOLID   GEOMETRY 


303.  Theorem.  —  If  a  straight  line  is  perpendicular  to  each 
of  two  intersecting  straight  lines  at  their  point  of  intersection^ 
it  is  perpendicular  to  the  plane  determined  hy  them. 


Hypothesis.  Lines  QJ)  and  EF  intersect  at  0,  and  deter- 
mine plane  MN\  AB  ±  CD  and  AB  ±  EF  at  0. 

Conclusion.     Line  AB  is  perpendicular  to  plane  MN. 

Proof.  1.  In  plane  MN,  CD  and  EF  intersect  at  0; 
also  AB  ±  CD  and  AB  ±  EF  at  0.  Hyp. 

2.  Draw  OK,  any  other  straight  line  through  0  in  MN. 
Draw  any  straight  line  intersecting  CD  at  (7,  EF  at  H,  and 
OK  At  K  On  AB,  take  B  on  the  opposite  side  of  0  from 
A  so  that  OB  =  AO.     Draw  AG,  AH,  AK,  Ba,  BE,  ^K 

3.  Then  Aa=Ba  and  AH=^  BE.  §  105 

4.  GE  is  a  common  side  of  A  AGE  and  A  BGE. 

5.  ..AAGE^ABGE.  §76 

6.  .'.  ZAGE==  ZBGE.  Def.  congruence 

7.  In  A  J.  (t^ and  A  BGK,  GK is  a  common  side. 

8.  .'.AAGK^ABGK  §  63 

9.  .'.AK=BK.  Def.  congruence 

10.  .'.AB±OK.  §  107 

11.  Hence,  since  OK  is  any  other  line  than  CD  and  EF 
drawn  through  0  in  MN,  AB  is  perpendicular  to  the 
plane  MN.  §  302 

Write  the  proof  in  full  without  referring  to  the  book. 


LINES   AND   PLANES   IN   SPACE 


287 


304.  Theorem.  —  Through  a  given  point  in  a  given  straight 
line^  one  plane  and  only  one  plane  can  he  drawn  perpendicular 
to  the  line. 


B 


M/ " ;      Mrr===^=^ 


B 


Hypothesis.  AB  is  a  given  straight  line,  and  0  is  a  given 
point  on  AB, 

Conclusion.  Through  0,  one  plane  and  only  one  plane  can 
be  drawn  perpendicular  to  AB, 

Proof.  1.  AB  is  a  given  straight  line,  and  (9  is  a  given 
point  on  AB,  Hyp. 

2.  Two  straight  lines,  OQ  and  OB  (figure  at  left),  can 
be  drawn  perpendicular  to  AB  at  0,  §  294 

3.  0(7  and  OB  determine  a  plane  MN.  §  298 

4.  Plane  MN 1.  line  AB,  §  303 

5.  Suppose  that  MN"is  not  the  only  plane  through  0  per- 
pendicular to  AB,  and  that  PQ  (figure  at  right)  is  another 
plane  through  0  perpendicular  to  AB. 

6.  Then  a  plane  can  be  drawn  containing  AB  and  inter- 
secting planes  iKfZVand  PQ  in  two  straight  lines  OF  Siud  OE, 
respectively.  §  301 

7.  OF  ±  AB  and  OE  i.  AB.  §  302 

8.  But  this  is  impossible.  §  53 

9.  .♦.  the  supposition  is  false,  and  hence  MN  is  the  only 
plane  through  0,  perpendicular  to  AB. 

Draw  figures,  and  write  the  proof  in  full  without  consulting 
the  book'. 


288 


SOLID   GEOMETRY 


305.  Theorem.  —  Through  a  given  point  without  a  given 
straight  line,  one  plane  and  only  one  plane  can  he  drawn  per- 
pendicular to  the  line. 


B 


I^^\ 


N 


A 

— ^ 

yr-F 

-       n\\ 

1       ' 

"^  \ 

pl--2 

Suggestions.  If  AB  (figure  at  left)  is  tl:ie  line  and  0  the 
point,  0  and  AB  determine  a  plane,  in  which  OQ  can  be 
drawn  ±  AB,  Then  another  line  CD  can  be  drawn  ±  AB, 
0(7  and  CD  determine  a  plane  MN  L  AB,  Show  (figure  at 
right)  by  using  §  54  that  no  other  plane  PQ  can  be  drawn 
through  0  J_  line  AB. 

306.  Theorem.  —  Through  a  given  point  in  a  given  plane, 
one  line  and  only  one  line  can  he  drawn  perpendicular  to  the 
plane. 

R 


p 

E 

M/c 

A\0 

0 

JO 


!\ 


A^ 


Suggestions.  If  0  (figure  at  left)  is  a  point  in  plane 
MN,  straight  line  AB  can  be  drawn  through  0  in  MN \  then 
plane  FQ  can  be  drawn  through  01.  AB,  intersecting  MN 
in  CD ;  then  line  OE  can  be  drawn  in  P^  ±  CD.  Then 
0^±  plane  ilOT. 

Show  (figure  at  right)  by  using  §  53  that  no  other  line  Oi^ 
can  be  drawn  through  0  ±  plane  MN, 


LINES   AND   PLANES   IN   SPACE 


289 


307.  Theorem.  —  Through  a  given  point  without  a  given  plane, 
one  line  and  only  one  line  can  he  drawn  perpendicular  to  the 
plane. 

P 


0 

K 

/ 

G 

.  'H-' 

\ 

7^\ 

H 

N 
Q 

Ok 


ML 


N 


Hypothesis.     MNis  a  plane  and  0  a  point  not  in  MN. 
Conclusion.     Through  (9  one  line  and  only  one  line  can  be 
drawn  perpendicular  to  MN. 

Proof.     1.    MJSf  is  a  plane  and  0  a  point  not  in  MN,  Hyp. 

2.  Let  AB  be  any  straight  line  in  MN  (figure  at  left). 

3.  A  plane  FQ  can  be  drawn  through  0  ±  AB,        §  305 

4.  FQ  intersects  MJVin  a  straight  line  CB,  and  line  AB 
at  a  point  U.  §  301 

5.  Draw  OG  in  FQ±  OB.  Let  GF  be  any  straight  line 
drawn  in  iHfiV,  meeting  AB  at  F.  Produce  OG  through  G 
to  IT,  making  GH^OG.     Draw  OE,  OF,  HE,  HF. 

6.  AB  ±  OE  siud  AB  ±  HE.  §302 

7.  .-.  Z  OFF  and  Z  HEF  are  rt.  A.  Dei.  ± 

8.  OE^HE.  §105 

9.  In  A  OEF  and  A  ^^^,  EF  is  a  common  side. 

10.  .-.A  OEF  ^  A  HEF.  §  64 

11.  .♦.  OF  =  HF.  Def.  congruence 

12.  .-.  GFl.  OG.  §  107 

13.  .-.  OG  i^  perpendicular  to  plane  MN.  §  303 
The  proof  that  0(7  is  the  only  line  that  can  be  drawn 

through   0  perpendicular   to   MN  is   left   to   the   student. 
Use  the  figure  at  the  right.     Employ  §  54. 


290 


SOLID   GEOMETRY 


EXERCISES 

1.  Explain  how  to  test  by  means  of  a  carpenter's  squarvi  whether  or 
not  a  post  on  a  level  surface  is  vertical. 

2.  Establish  a  line  perpendicular  to  the  top  of  the  table  by  using 
two  rectangular  pieces  of  cardboard  or  two  books. 

3.  How  can  a  carpenter  determine  if  a  floor  is  level  by  means  of  a 
plumb  line  and  a  steel  square  ? 

4.  Explain  how  a  carpenter  could  set  a  timber  perpendicular  to  the 
floor  by  using  only  a  ten-foot  pole  notched  at  the  six-foot  and  eight-foot 
points. 

5.  Show  how  by  §  303  a  carpenter  is  enabled  to 
saw  a  piece  of  lumber  squarely  in  two. 

6.  Prove  that  any  point  in  the  plane  perpen- 
dicular to  a  given  line-segment  at  its  middle  point  is 
equidistant  from  the  ends  of  the  line-segment. 

7.  Prove  that  any  point  of  space  equidistant  from 
the  ends  of  a  given  line-segment  is  in  the  plane  per- 
pendicular to  the  line-segment  at  its  middle  point. 

8.  Prove  that  if  a  plane  is  perpendicular  to  a  line-segment  at  its 
middle  point,  any  point  not  in  the  plane  is  unequally  distant  from  the 
ends  of  the  line-segment. 

9.  What  is  the  locus  of  points  in  space  equidistant  from  two  given 
points  ? 

10.  Find  the  locus  of  points  in  a  given  plane  that  are  equidistant  from 
any  two  given  points  not  in  that  plane. 

11.  A  ray  of  light  that  starts  from  a  given  point  B  and  is  reflected 
by  a  mirror  to  a  given  point  A  travels  along  the  shortest  possible  path. 
Find  the  point  of  the  mirror  from  which  the 
ray  is  reflected. 

Suggestion.  —  Draw  a  perpendicular  from 
one  of  the  given  points  to  the  plane  of  the 
mirror,  and  extend  it  to  a  point  an  equal  dis- 
tance on  the  other  side  of  the  plane  of  the 
mirror.  Join  this  point  to  the  other  given 
point.  Prove  that  the  point  at  which  this 
second  line  meets  the  surface  of  the  mirror  is  the  required  point. 


LINES   AND   PLANES   IN   SPACE  291 

308.  Theorem.  —  All  lines  perpendicular  to  a  given  line  at  a 
given  point  lie  in  the  plane  perpendicular  to  the  given  line  at 
the  given  point. 

A 

n \N 


Hypothesis.  Plane  MN  is  perpendicular  to  line  AB  at  0  ; 
OOis  any  line  perpendicular  to  AB  at  0. 

Conclusion.     OClies  in  MN. 

Suggestions.  Suppose  that  00  does  not  lie  in  MN.  Then 
the  plane  determined  by  AB  and  00  must  intersect  MN  in 
another  line  OB.     Show  that  this  violates  §  53. 

Write  the  proof  in  full. 

EXERCISES 

1.  If  a  spoke  of  a  wheel  is  perpendicular  to  the  axle  upon  which  it 
turns,  it  describes  a  plane  in  its  rotation. 

2.  The  locus  of  all  lines  perpendicular  to  a  given  line  at  a  given  point 
is  the  plane  perjifendicular  to  the  given  line  at  the  given  point. 

3.  If  a  plane  is  perpendicular  to  a  given  line,  any  line  perpendicular 
to  the  given  line  through  any  point  of  the  given  plane  must  lie  in  the 
given  plane. 

309.  Theorem.  —  77ie  perpendicular  to  a  plane  from  an  ex- 
ternal point  is  the  shortest  line-segment  that  can  be  drawn  to  the 
plane  from  the  point. 

The  proof  is  left  to  the  student.     Write  the  proof  in  full. 

The  perpendicular  line-segment  drawn  from  a  given  point 
to  a  given  plane  is  called  the  distance  from  the  point  to  the 
plane. 


292  SOLID   GEOMETRY 

310.  Theorem.  —  (1)  Oblique  line-segments  drawn  from  a 
point  to  a  plane^  meeting  the  plane  at  equal  distances  from  the 
foot  of  the  perpendicular  from  the  pointy  are  equal ;  (2)  of  two 
oblique  line-segments  meeting  the  plane  at  unequal  distances 
from  the  foot  of  the  perpendicular^  the  more  remote  is  the 
greater. 


^R 


Hypothesis.  Line  AB  is  perpendicular  to  plane  MN"^  meet- 
ing MN  at  B  ;  oblique  line-segments  from  A  meet  MN  at  (7, 
D,  B  ;  BI)=^BCdnd  BB  >  BO. 

Conclusion.     AB  =  AO,  and  AB  >  A  O. 

Suggestions.  It  may  be  proved  that  AI)  =  AO  if  it  is 
proved  that  A  ABB  ^AABO. 

For  i^roYing  AB>AO,  on  BB  mark  o^BF=BO,  and 
draw  AF.  Point  F  lies  between  B  and  B.  Why  ?  It  can 
be  proved  that  AB  >  AO  il  it  is  shown  that  AO  =  AF  and 
AB  >  AF.     But  AB  >  AF  ii  Z.  AFB  >  Z  ASF,  etc. 

Write  the  proof  in  full. 

311.  Theorem.  —  (1)  Bqual  oblique  line-segments  drawn  from 
a  point  to  a  plane  meet  the  plane  at  equal  distances  from  the 
foot  of  the  perpendicular  from  the  point;  (2)  of  two  unequal 
line-segments^  the  greater  meets  the  plane  at  the  greater  distance 
from  the  foot  of  the  perpendicular. 

Suggestions.  This  theorem,  which  is  the  converse  of  the 
theorem  in  §  810,  may  be  proved  easily  by  the  indirect 
method,  making  use  of  §  310. 

The  proof  is  left  to  the  student.     Write  the  proof  in  full, 


LINES  AND  PLANES  IN  SPACE 


293 


EXERCISES 

1.  If  a  derrick  stands  vertically  on  level  ground,  guy  ropes  reaching 
from  the  top  of  the  derrick  to  stakes  in  the  ground  at  equal  distances 
from  the  foot  of  the  derrick  are  equal. 

2.  If  a  line-segment  AB  is  perpendicular  to  a  plane  at  B,  it  subtends 
equal  angles  at  all  points  of  a  circle  lying  in  the  plane  and  having  B  as 
center. 

3.  If  the  line-segment  AB  in  perpendicular  to  a  plane,  and  intersects 
the  plane  at  a  point  C  on  AB  produced,  AB  subtends  equal  angles  at  all 
points  of  a  circle  lying  in  the  plane  and  having  C  as  center. 

4.  If  the  line-segment  AB  is  perpendicular  to  a  plane,  and  intersects 
the  plane  at  a  point  C  between  A  and  B,  AB  subtends  equal  angles  at 
all  points  of  a  circle  lying  in  the  plane  and  having  C  as  center. 

5.  If  a  line-segment  AB  is  ]-)erpendicular  to  a  plane  at  B,  and  a  circle 
is  drawn  in  the  plane,  with  center  at  J5, 
then  the  angle  subtended  by  AB  B,t  a,  point 
of  the  plane  within  the  circle  is  greater,  and 
at  a  point  without  the  circle  less,  than  the 
angle  subtended  at  a  point  on  the  circle. 

Suggestion.  —  Prove  ZAEB>ZACB 
and  ZADB<ZACB. 

6.  If  a  line-segment  AB  is  perpendicular  to  a  plane  at  B,  and  a  circle 
is  drawn  in  the  plane,  with  center  at  /i,  then  a  point  of  the  plane  is 
within,  on,  or  without  the  circle,  according  as  AB  subtends  at  that  point 
an  angle  greater  than,  equal  to,  or  less  than  the  angle  which  it  subtends 
at  a  point  on  the  circle. 

7.  A  ship  may  be  steered  past  a  region  of  danger  by  observing  the 
angle  of  elevation  at  the  ship  sub- 
tended by  a  landmark  A,  within  this 
region,  as  follows  :  A  map  contains  a 
circle  with  the  foot  of  A  as  center, 
and  large  enough  to  inclose  the  region 

of  danger.     The  size  of  the  angle  m _ 

which  the  landmark  A  subtends  at  a 

point  of  the  circle  is  noted  on  the  map.  The  course  of  the  ship  is  so  di 
rected  that  the  angle  of  elevation  of  A  as  observed  from  the  ship  from 
time  to  time  never  becomes  greater  than  angle  m.  Prove  that  the  ship 
does  not  enter  the  region  of  danger. 


294  SOLID   GEOMETRY 

8.  If  the  ceiling  of  a  room  is  ten  feet  high,  how  can  a  point  of  the 
floor  that  is  direct ly  under  a  given  point  of  the  ceiling  be  located  by  using 
only  an  11-foot  pole,  chalk,  and  a  string? 

9.  What  is  the  locus  of  the  feet  of  equal  line-segments  drawn  to  a 
plane  from  a  point  in  a  perpendicular  to  the  plane  ?    Prove  it. 

10.  The  locus  of  points  each  of  which  is  equidistant  from  all  points  of 
a  circle  is  a  straight  line  through  the  c&nter  and  perpendicular  to  the 
plane  of  the  circle. 

11.  If,  from  the  foot  of  a  perpendic- 
ular to  a  plane,  a  line  is  drawn  perpen- 
dicular to  any  given  line  in  the  plane, 
the  line  which  joins  the  point  of  inter- 
section to  any  point  in  the  perpendicular 
to  the  plane  is  perpendicular  to  the 
given  line  in  the  plane. 

Suggestions.  —  Let  AB  he  perpendicular  to  plane  M"iV;  let  CD  be 
any  line  in  MN;  let  BE  ±  CD.     It  is  required  to  prove  AE  ±  CD. 

Mark  off  CE  =  DE,  and  draw  BC,  BD,  AC,  AD. 

It  can  be  proved  that  AE  ±  CD  if  it  is  first  proved  that  AC  =  AD. 
It  can  be  proved  that  AC  =  AD  if  it  is  first  proved  that  AABC ^ 
A  ABD,  etc. 

12.  Prove  the  converse  of  Exercise  11,  i.e.,  if  ^5  ±  plane  ikfiV  and 
AE±CD,  then  BE  ±  CD. 

Suggestions.  —  Mark  ofi  CE  =  DE  and  draw  the  auxiliary  lines 
AC,  BC,  AD,  and  BD  as  in  the  proof  of  Exercise  11. 

It  may  be  proved  that  BE  ±  CD  if  what  is  proved  first?  How  may 
the  latter  be  proved  ? 

13.  If  a  given  line  is  perpendicular  to  a  given  plane,  all  of  the  lines 
which  are  perpendicular  to  a  line  of  the  plane  from  points  of  the  given 
line  are  concurrent. 

Suggestion.  —  In  the  figure  of  Exercise  11,  A  E  intersects  CD  at  what 
special  point? 

312.  Parallel  lines  and  planes.  —  A  straight  line  and  a  plane 
which  do  not  intersect  are  said  to  be  parallel. 

Two  planes  which  do  not  intersect  are  called  parallel  planes. 


LINES  AND   PLANES   IN   SPACE  295 

313.    Theorem.  —  If  a  straight  line  not  in  a  given  plane  is 
parallel  to  a  line  in  the  plane^  it  is  parallel  to  the  plane. 


C 


B 

— V^ 


Hypothesis.  Line  CD  lies  in  plane  MN  \  line  AB  II  (7Z),  and 
AB  is  not  in  MN. 

Conclusion.     AB  is  parallel  to  plane  MN. 

Proof.  1.  Line  CD  lies  in  plane  MN\  line  AB  II  OD,  and 
AB  is  not  in  MN.  Hyp. 

2.  .-.  AB  and  QD  are  in  a  plane.  §  299 

3.  This  plane  intersects  plane  MN  in  CD,  because  CD  is 
in  both  planes. 

4.  Hence,  if  AB  met  MN^  the  point  of  intersection  would 
be  on  (7i>,  because  it  would  be  in  both  planes. 

5.  But  AB  cannot  meet  CD.  Def.  II 

6.  .-.  AB  cannot  meet  MN. 

7.  .-.  AB  is  parallel  to  MN.  §  312 

EXERCISES 

1.  State  the  converse  of  the  theorem  iu  §  313.  Is  it  a  true  theorem? 
Illustrate. 

2.  Through  a  given  point  without  a  given  straight  line,  an  unlimited 
number  of  planes  can  be  passed  which  are  parallel  to  the  given  line. 

3.  If  two  straight  lines  are  not  in  the  same  plane,  a  plane  can  be  drawn 
containing  one  of  them  and  parallel  to  the  other. 

4.  Through  a  given  point  without  a  given  plane,  any  number  of  lines 
can  be  drawn  parallel  to  the  plane. 

5.  If  two  intersecting  planes  are  drawn  through  two  parallel  lines,  theii 
line  of  intersection  is  parallel  to  each  of  the  lines. 


296 


SOLID   GEOMETRY 


314.  Theorem.  —  If  a  straight  line  is  parallel  to  a  plane^  the 
intersection  of  this  plane  and  any  plane  containing  the  given  line 
is  parallel  to  the  given  line. 


Mz 


D 


.N 


Hypothesis.     Line  AB  is  parallel  to  plane  MN \  plane  AD 
contains  AB^  and  intersects  plane  MN  in  line  QD. 

Conclusion.      OD  II  AB. 

Suggestion.     If  AB  and  QD  intersect,  AB  must  intersect 

plane  MN^  which  is  impossible. 
Write  the  proof  in  full. 


315.    Theorem.  — If  a  plan^  ^dtersects  two  parallel  planes^  the 
lines  of  intersection  are  parallel. 


M 

P^<^7\. 

y*^ 

^^\ 

^o 

.y            ^ 

^/^/' 

c 


N 


Hypothesis.     Plane  P§  is  parallel  to  plane  RS ;  plane  MN 
intersects  plane  PQ  in  AB  and  plane  ^aS'  in  CD. 

Conclusion.     AB  II  CD. 

Suggestion.     AB  and  CD  lie  in  the  same  plane  and  cannot 
intersect.     Why  ? 

Write  the  proof  in  full. 


LINES  AND   PLANES   IN   SPACE  297 

EXERCISES 

1.  Segments  of  parallel  lines  which  are  cut  off  by  two  parallel  planes 
are  equal. 

2.  If  a  line  is  parallel  to  a  plane,  a  plane  can  be  drawn  containing  the 
given  line  and  parallel  to  the  given  plane. 

3.  If  a  straight  line  and  plane  are  parallel,  any  straight  line  drawn 
through  a  point  of  the  plane  and  parallel  to  the  given  line  lies  wholly  in 
the  given  plane. 

Suggestions.  —  The  two  parallel  lines  determine  a  plane,  which  in- 
tersects the  given  plane  in  a  line  through  the  given  point. 

4.  If  a  line  is  parallel  to  each  of  two  intersecting  planes,  it  is  parallel 
to  their  line  of  intersection. 

5.  If  a  straight  line  is  parallel  to  a  plane,  any  straight  line  which  is 
parallel  to  the  given  line  and  not  in  the  given  plane  is  also  parallel  to  the 
given  plane. 

Suggestion.  —  The  two  parallel  lines  determine  a  plane.  Consider 
the  two  cases  when  this  plane  is  parallel  to  the  given  plane  and  when  it 
intersects  it.  In  the  latter  case,  what  is  known  about  the  line  of  in- 
tersection ? 

6.  If  each  of  two  intersecting  straight  lines  is  parallel  to  a  given 
plane,  the  plane  determined  by  these  lines  is  parallel  to  the  given  plane. 

Suggestion.  —  Use  indirect  proof.     Apply  §  314,  then  §  46. 

7.  If  two  lines  of  one  plane  are  paral- 
lel respectively  to  two  intersecting  lines 
of  another  plane,  the  planes  are  parallel. 

Suggestions.  — If  ^5  II  ^F and  CD  II 
GH,  what  relation  has  plane  MN  to  EF 
and  to  GH'i    Now  apply  Exercise  6. 

8.  Through  a  given  point  a  plane  can 

be  passed  parallel  to  any  two  given  straight  lines  in  space. 

Suggestion.  —  Through  the  given  point,  draw  lines  parallel  to  the 
given  lines. 

9.  If  a  straight  line  is  parallel  to  each  of  two  given  planes,  the  lines  of 
intersection  which  any  plane  passing  through  the  given  line  makes  with 
the  given  planes  are  parallel. 


298  SOLID   GEOMETRY 

316.    Theorem.  —  Two  straight  lines  which  are  parallel  to  a 
third  straight  line  not  in  their  plane  are  parallel  to  each  other. 


GF 


Hypothesis.     CD  II  AB  and  EF II  AB. 
Conclusion.      OB  II  EF. 

Proof.     1.    OB  II  AB  and  EF  II  AB,  •       Hyp. 

2.  AB  and  OB  determine  a  plane  AD^  and  AB  and  EF 
determine  a  plane  AF.  §  299 

3.  OD  and  point  E  determine  a  plane  OGr.  §  296 

4.  Planes  OG-  and  ^^  intersect  in  a  line  J57(^.  §  301 

5.  (7i)  II  plane  ^^.  §313 

6.  .-.  OB  II  ^a.  §  314 

7.  AB  II  plane  Oa,  §  313 

8.  :•.  AB  II  J5;(7.  §  314 

9.  .-.  Ea  and  jE:^  coincide.  §  46 
10.  .-.  (7i>  II  EF.  Step  6 
Write  out  the  proof  without  the  book. 

EXERCISES 

1.  If  a  straight  line  and  plane  are  parallel,  the  lines  of  intersection  of 
th^  given  plane  and  all  planes  which  contain  the  given  line  are  parallel. 

2.  In  a  rectangular  room  the  intersection  of  the  ceiling  and  a  wall  is 
parallel  to  the  intersection  of  the  floor  and  the  opposite  wall. 

3.  In  a  quadrilateral  whose  sides  are  not  all  in  the  same  plane  (called 
a  gauche  quadrilateral),  the  line-segments  which  join  the  middle  points 
of  the  adjacent  sides  form  a  parallelogram. 


LINES  AND   PLANES   IN   SPACE  299 

317.  Theorem.  —  If  two  angles  in  different  planes  have  their 
sides  parallel  each  to  each  and  extending  in  the  same  direction 
from  the  vertices^  the  angles  are  equal  and  their  planes  are 
parallel. 


Hypothesis.  Z  ABO  is  in  plane  MN,  and  Z  DEF  is  in 
plane  PQ  ;  BA  II  ED  and  BQ II  EF,  and  the  parallel  lines  ex- 
tend in  the  same  directions  from  the  vertices. 

Conclusion.     Z.  ABO  =  Z.  DEF  and  MN II  P Q, 

Proof.     1.    BA  WED  and  BOW  EF,  Hyp. 

2.  Mark  off  on  the  sides  of  the  angles  BA  =  ED   and 
BO=EF.     Draw  BE,  AD.  OF,  AO,  DF. 

3.  Then  EDAB  and  EFOB  are  parallelograms.  §  90 

4.  .-.  AD  II  BE  and  OF  H  BE.  Dei  O 

5.  .-.  AD  11  OF.  §  316 

6.  Also  AD  =  BE  and  (TF  =  BE,  §  82 

7.  .'.AD^OF.  Ax.  I 

8.  .*.  ADFO  =  CJ,  §  90 

9.  .'.AO=DF.  §82 

10.  .'.AABO^ADEF.  §76 

11.  .-.  Z  ^5(7  =  Z  i)^jP.  Def.  congruence 

12.  Now  BA  II  plane  P§  and  BOW  plane  P^.  §  313 

13.  If  MZVand  P§  intersected,  their  line  of  intersection 
would  meet  either  BA  or  B  0  or  both.  §  46 

14.  But  this  is  impossible.  §  312 

15.  .-.  MN  and  PQ  do  not  intersect ;  and  therefore 
MNWPQ.  §312 


300 


SOLID   GEOMETRY 


318.    Theorem.  — A  straight  line  perpendicular  to  one  of  two 
parallel  planes  is  perpendicular  to  the  other  also. 


m/ 

-------£p  \ 

y  ---^ 

E           N^- 

Hypothesis.     Plane  MNW  plane  PQ  ;  line  AB  ±  plane  PQ. 
Conclusion.     Line  AB  X  plane  MN. 

Proof.     1.    Plane  MN  II  plane  PQ  ;    and  line  AB  ±  plane 
PQ.  Hyp. 

2.  Let  two  planes  containing  AB  intersect  ilifiVin  ^(7  and 
AT).,  respectively,  and  PQ'\w  BE  and  BP.,  respectively. 

3.  Then  AOWBE  and  AB  11  BF.  §  315 

4.  But  AB  X  BE  and  AB  ±  BF.  §  302 

5.  r.ABA.AO?^ndiAB^^AD.  §34 

6.  .-.  line  ^^  J-  plane  MN.  §  302 
Draw  a  figure  and  write  the  proof  without  the  book. 

319.    Theorem.  —  Two  planes  which  are  perpendicular  to  the 
same  straight  line  are  parallel. 


./ 

A                       \^ 

./ 

B                 M( 

Hypothesis.  Plane  il!fiV±  line  J.5  at  A\  plane  P§±line 
AB  at  ^. 

Conclusion.     MNW  PQ. 

Suggestion.  Suppose  MN  and  PQ  not  parallel,  and  from 
any  point  of  their  line  of  intersection,  lines  drawn  to  A  and  B. 

Write  the  proof  in  full. 


LINES   AND   PLANES   IN   SPACE 


301 


320.    Theorem.  —  If  one  of  two  parallel  lines  is  perpendicular 
to  a  plane,  the  other  is  also  perpendicular  to  the  plane, 

C 


Mi 


B 


D 


.N 


Hypothesis.  Lines  AB  and  CI)  meet  plane  MJV  at  B  and 
Z>,  respectively  ;  ^5  ±  plane  MN;  AB  II  CD. 

Conclusion.     OB  ±  plane  MK 

Proof.     1.    AB  ±  plane  MN  and  AB  II  OB.  Hyp. 

2.  In  ifiV^  draw  any  line  DF  from  i),  and  draw  BU  II  BF. 
Then  Z  ABE  =  Z  OBF.  §  317 

But  J.5  ±  BE.  §  302 

.-.  Z^^^  =  rt.  Z.  Def.  ± 

.'.  Z  CDF  =rLZ,  Ax.  I 

Hence,  since  BF  is  awy  line  in  MN'  through  2>,  Ci)  J- 


plane  MN, 


§  302 


321.    Theorem.  —  Two  lines  perpendicular  to  the  same  plane 
are  parallel  to  each  other. 


mL 


B' 


lE 


D 


.N 


Hypothesis.  Line  ^5  _L  plane  MN-,  line  (7i)±  plane  MN. 

Conclusion.  AB  11  CB. 

Suggestion.  Suppose  AB  not  parallel  to  CB,  but  EB  II  CB. 

Apply  §  320.  Write  out  the  complete  proof. 

322.  Distance  between  parallel  planes.  —  A  line-segment  con- 
necting points  in  two  parallel  planes  and  perpendicular  to 
each  plane  (See  §  318)  is  called  the  distance  between  the  par- 
allel planes. 


302  SOLID  GEOMETRY 

323.    Theorem.  —  Two  parallel  planes  are  everywhere  equally) 
distant  . 


M^ 


C 


M 


P^ 


D 


vO 


Hypothesis.     Plane  MNW  plane  PQ, 

Conclusion.     MN  and  PQ  are  everywhere  equally  distant. 

Suggestion.  Let  AB  and  CD  be  any  two  line-segments  be- 
tween il^ZVand  PQ^  each  being  perpendicular  to  both  planes. 
Prove  AB  =  OB,     Write  the  proof  in  full. 

EXERCISES 

1.  If  three  equal  line-segments  which  are  not  in  one  plane  are  paral- 
lel, the  triangles  formed  by  joining  their  corresponding  end  points  are 
congruent  and  lie  in  parallel  planes. 

2.  Two  points  on  the  same  side  of  a  plane  and  equally  distant  from 
it  determine  a  line  that  is  parallel  to  the  plane. 

3.  Through  a  given  point  only  one  plane  can  be  passed  parallel  to  a 
given  plane. 

4.  Two  planes  each  parallel  to  a  third  plane  are  parallel  to  each  other. 

5.  A  straight  line  and  a  plane,  both  perpendicular  to  the  same  straight 
line,  are  parallel. 

6.  What  is  the  locus  of  points  equidistant  from  two  given  parallel 
planes  ?    Prove  it. 

7.  What  is  the  locus  of  points  equidistant  from  anj  two  given  points 
and  also  equidistant  from  two  given  parallel  planes  ?    Prove  it. 

8.  What  is  the  locus  of  points  at  a  given  distance  from  a  given  plane  ? 
Prove  it. 

9.  Find  a  point  equidistant  from  two  given  points,  equidistant  from 
two  given  parallel  planes,  and  at  a  given  distance  from  a  third  given 
plane. 

10.  If  two  angles  in  different  planes  have  their  sides  parallel  each  to 
each,  one  pair  of  parallel  sides  extending  in  the  same  direction  and  the 
other  pair  in  opposite  directions  from  the  vertices,  the  angles  are_ supple- 
mentary. 


LINES  AND   PLANES  IN   SPACE  303 

324.  Theorem.  —  If  two  straight  lines  intersect  three  parallel 
planes^  their  corresponding  segments  cut  off  hy  the  planes  are 
proportional. 


X 


Hypothesis.     Planes  MN,  PQ,  and  MS  are  parallel,  and  cut 

off  segments  AB  and  BO  from  line  AC,  and  corresponding 

segments  BU  and  UF  from  line  BF, 

n      .    '        AB     BE 
Conclusion.    — —  =  — —  • 
BO     EF 

Suggestions.     Draw  AF,  intersecting  PQ  at  G-^  and  draw 

BGi,  OF^  AB,  and  GrB.     The  conclusion  may  be  established 

if  it  is  first  proved  that  — —  and  — — -  are  each  equal  to  -;r— -• 

BO         EF  CrF 

This  may  be  proved  if  it  is  first  proved  that  BQ-  II  OF,  etc. 

Write  the  proof  in  full. 

EXERCISES 

1.  A  line  meets  three  parallel  planes  in  the  points  A,  B,  and  C,  re- 
spectively. A  second  line  meets  the  planes  in  the  corresponding  points 
D,  E,  and  F,  respectively.  AB  =  6  in.,  BC  =  8  in.,  and  DF=  18  in. 
Find  the  lengths  of  DE  and  EF. 

2.  The  distances  between  four  shelves  are  6  in.,  8  in.,  and  10  in.,  re- 
spectively. A  diagonal  brace  36  in.  long  reaches  from  the  top  shelf  to 
the  bottom  shelf.  Find  the  segments  of  the  brace  between  the  shelves, 
making  no  allowance  for  the  thickness  of  the  shelves. 

3.  If  three  parallel  planes  intercept  equal  segments  on  one  straight 
line,  they  intercept  equal  segments  on  any  other  straight  line  which  they 
intersect. 

4.  If  two  straight  lines  intersect  any  number  of  parallel  planes,  their 
corresponding  segments  are  proportional. 


D 


304  SOLID   GEOMETRY 

325.  Diedral  angles.  —  When  two  planes  intersect,  the  line 
of  intersection  divides  each  plane  into  two  parts.  One  such 
part  of  one  plane  and  one  such  part  of  the  other  plane  to- 
gether form  a  figure  called  a  diedral  angle. 

The  parts  of  planes  forming  a  diedral  angle  are  called  the 
faces  of  the  diedral  angle,  and  the  line  of  intersection  of  the 
planes  is  called  the  edge  of  the  diedral 
angle. 

Thus,  in  the  figure,  the  planes  ylC  and  BD 
are  the  faces,  and  the  line  AB  is,  the  edge  of     A 
the  diedral  angle. 

A  diedral  angle  is  named  by  naming 
a  point  in  one  face,  then  the  edge,  then  a  point  in  the  other 
face.     When  confusion  would  not  result,  a  diedral  angle  may 
be  named  by  merely  naming  its  edge. 

Thus,  the  diedral  angle  above  is  named  "angle  C-AB-D^'  or  merely 
"  angle  AB^ 

326.  The  plane  angle  of  a  diedral  angle.  —  The  angle  formed 
by  straight  lines  drawn  in  the  two  faces  of  a  diedral  angle, 
perpendicular  to  the  edge  at  the  same 
point  of  the  edge,  is  called  the  plane  angle 
of  the  diedral  angle. 


Thus,  if  PE ±AB  and   PE  lies  in  face  BC, 
and  if   PF±AB   and  PF  lies  in   face  AD,   of 
diedral  angle  C-AB-D,  then  Z  EPF  is  the  plane  angle  of  diedral  angle 
C-AB-D. 

It  is  evident  from  §  317  that  the  plane  angle  of  a  diedral 
angle  has  the  same  size  at  whatever  point  of  the  edge  its 
vertex  is  taken. 

Let  the  student  draw  a  figure  representing  any  two  posi- 
tions of  the  plane  angle  of  a  diedral  angle,  and  reason  out 
this  truth. 


LINES  AND  PLANES  IN   SPACE 


305 


327.   Theorem.  —  Two  diedral  angles  are  equal  if  their  plane 
angles  are  equal. 


Hypothesis.  /.  MON  and  Z  PQB  are  plane  angles  of 
diedral  angles  C-AB-B  and  G-EF-H,  respectively;  and 
Z  MO]Sr=  Z  PQR. 

Conclusion.     Angle  C-AB-D  =  angle  G-EF-H. 

Proof.  1.  Z  MONwmX  Z  PQR  are  plane  angles  of  diedral 
angles  C-AB^D  and  G-EF-ff,  respectively ;  and  Z  MOJSf 
=  Z  PQR.  Hyp. 

2.  .-.  /LMON  can  be  superposed  on  Z  P§i2  so  that  they 
coincide.     Let  it  be  so  placed.  §  13 

3.  AB  _L  OM,  AB  X  Oi\r;  also  EF  JL  QP,  EF±  QR. 

§  326 

4.  .-.^^Xplaneof  Zil[fOiV;and^PJ-planeof   ZP^jR. 

§  303 

5.  But  plane  of  Z  il[fOiVand  plane  of  Z  P^i2  coincide. 

§  298 

6.  .-.  AB  and  ^P  coincide.  §  306 

7.  .-.  the  faces  of  angle  C-AB-B  and  the  faces  of  angle 
G-EF-H  coincide.  §  298 

8.  .-.  angle  C-AB-B  =  angle  G-EF-H^  because  they 
coincide  throughout. 

328.  Theorem.  ^-If  two  diedral  angles  are  equals  their  plane 
angles  are  equal. 

The  proof  of  this  theorem  is  left  to  the  student.  Use 
superposition,  as  in  §  327.     Write  the  proof  in  full. 


306  SOLID   GEOMETRY 

329.  Special  diedral  angles. —  It  follows  from  §327  and 
§  328  that  the  plane  angle  of  a  diedral  angle  may  be  taken 
as  the  measure  of  the  diedral  angle. 

A  diedral  angle  is  called  a  right,  acute,  or  obtuse  diedral 
angle  according  as  its  plane  angle  is  right,  acute,  or  obtuse. 

Two  diedral  angles  are  called 
adjacent,  vertical,  complementary, 
or  supplementary,  according  as 
their  plane  angles  are  adjacent, 
vertical,  etc. 

If  two  planes  are  intersected 
by  a  third  plane,  the  terms  cor- 
responding diedral  angles^  etc.,  are  applied  to  the  pairs  of 
diedral  angles  in  the  same  sense  that  the  corresponding  terms 
are  applied  to  plane  angles. 

EXERCISES 

1.  Name  a  pair  of  vertical  diedral  angles  in  the  figure  of  §  329. 
Name  a  pair  of  adjacent  diedral  angles.  Name  a  pair  of  supplementary 
diedral  angles. 

2.  Draw  a  figure  representing  two  complementary  diedral  angles. 

3.  Prove  that  any  two  vertical  diedral  angles  are  equal. 

4.  Prove  that  any  two  right  diedral  angles  are  equal. 

5.  Prove  that  diedral  angles  which  are  complenjents  of  the  same 
diedral  angle  or  equal  diedral  angles  are  equal. 

6.  Prove  that  diedral  angles  which  are  supplements  of  the  same 
diedral  angle  or  equal  diedral  angles  are  equal. 

Prove  that  if  two  parallel  planes  are  cut  by  a  third  plane : 

7.  The  corresponding  diedral  angles  formed  are  equal. 

8.  The  alternate  interior  diedral  angles  formed  are  equal. 

9.  The  alternate  exterior  diedral  angles  formed  are  equal. 

10.   The  consecutive  interior  diedral  angles  formed  are  supplementary. 

330.  Perpendicular  planes.  —  Two  intersecting  planes  which 
form  a  right  diedral  angle  are  called  perpendicular  planes. 


LINES  AND   PLANES  IN   SPACE  307 

331.  Theorem.  — If  a  straight  line  is  perpendicular  to  a 
plane^  any  "plane  which  contains  the  line  is  perpendicular  to 
the  plane. 


I  ID 

m/  ^  "^     / 

c 


Hypothesis.  Line  AB  ±  plane  MN  at  A  ;  plane  PQ  con- 
tains AB. 

Conclusion.     Plane  PQ  1.  plane  MN. 

Proof.  1.  Line  ^^  ±  plane  MN  ?it  A;  plane  PQ  con- 
tains AB.  Hyp. 

2.  .-.  PQ  and  ife?iV  intersect  in  a  straight  line  OB.      §  301 

3.  Draw  AU  ±  OB  in  plane  MN 

4.  AB±CB.  §  302 

5.  Then  Z  UAB  is  the  plane  angle  of  diedral  angle 
N-CB-Q.  .        §  326 

6.  But  AB±  AH.  §  302 

7.  .-.  Z  BAB  =  rt.  Z.  Def.  ± 

8.  .-.  plane  PQ±  plane  MN.  §  330 

332.  Theorem.  —  A  straight  line  drawn  in  one  of  two  perpen- 
dicular planes,  perpendicular  to  their  line  of  intersection.,  is 
perpendicular  to  the  other  plane. 

Suggestion.  If  perpendicular  planes  MN  and  PQ  intersect 
in  line  (72>,  and  line  AB  1.  CB  and  lies  in  P§,  draw  line 
AE 1.  CB  in  plane  MN.  Then  the  analysis  is  as  follows: 
AB  ±  plane  MN  if  AB  J_  AE.  And  AB  1.AE  \i  Z  EAB 
=  rt.  Z.  And  Z  EAB  =  rt.  Z  if  Z  EAB  is  the  plane  angle 
of  diedral  angle  N-CB-Q  and  angle  N-CB-Q  is  a  right 
diedral  angle,  etc. 

Write  the  proof  in  full. 


308  SOLID   GEOMETRY 

333.  Corollary  l.  —  If  two  planes  are  perpendicular^  a  line 
perpendicular  to  one  of  them  at  any  point  in  their  intersection 
lies  in  the  other. 

Use  indirect  proof.     Apply  §  306  and  §  332. 

334.  Corollary  2.  —  If  two  planes  are  perpendicular,  a  line 
perpendicular  to  one  of  them  from  any  point  of  the  other  that 
is  not  in  the  intersection  lies  in  the  other  plane. 

Use  indirect  proof.     Apply  §  307  and  §  332. 

EXERCISES 

1.  In  sawing  a  board  in  two  with  a  polished  handsaw,  the  image 
of  the  board  is  seen  in  the  surface  of  the  saw.  When  the  image  of  the 
surface  of  the  board  and  the  surface 

of  the  board  itself  appear  to  form 
one  plane,  the  saw  is  cutting  the 
board  squarely  in  two,  i.e.  at  right 
angles.     Why? 

2.  The  edge  of  a  diedral  angle  is  perpendicular  to  the  plane  of  its 
plane  angle. 

3.  If  a  plane  is  perpendicular  to  the  intersection  of  two  planes,  it  is 
perpendicular  to  each  of  the  planes. 

4.  The  plane  of  the  plane  angle  of  a  diedral  angle  is  perpendicular 
to  the  faces  of  the  diedral  angle. 

5.  If  a  plane  is  perpendicular  to  a  line  in  another  plane,  it  is  perpen- 
dicular to  that  plane. 

6.  Through  a  given  straight  line  not  perpendicular  to  a  given  plane, 
one  and  only  one  plane  can  be  passed  perpendicular  to  the  given  plane. 

7.  If  three  lines  are  perpendicular  to  each  other  at  a  common  point, 
the  planes  of  the  lines  are  perpendicular  to  each  other. 

8.  If  a  line  and  a  plane  are  parallel,  any  plane  perpendicular  to  the 
line  is  also  perpendicular  to  the  plane. 

9.  If  three  or  more  planes  intersect  in  a  common  line,  the  lines  per- 
pendicular to  them  from  any  common  external  point  lie  in  one  plane. 


LINES  AND   PLANES  IN   SPACE 


309 


10.  If  two  planes  are  perpendicular  to  each  other,  a  line  perpendicular  to 
one  of  the  planes  and  not  contained  in  the  other  is  parallel  to  the  other. 

11.  If  a  line  is  parallel  to  one  plane  and  perpendicular  to  another,  the 
two  planes  are  perpendicular  to  each  other. 

12.  Between  two  straight  lines  not  in  the  same  plane,  one,  and  only- 
one  common  perpendicular  can  be  drawn. , 

SuGOESTiONS.  —  Let  AB  and  CD  be 
the  given  lines.  Pass  plane  MN  through 
CD  parallel  to  AB.  Through  AB,  pass 
a  plane  AE  perpendicular  to  MN,  inter- 
secting MN  in  PE  and  CD  at  P.  In  M. 
plane  AE  draw  PF ±PE.  Prove  PF  a  common  perpendicular  to  AB 
and  CD. 

Suppose  that  GQ  is  a  second  common  perpendicular,  draw  GH±PE, 
draw  QK  WAB,  and  show  that  an  absurdity  results. 

13.  The  common  perpendicular  between  two  lines  not  in  the  same 
plane  is  the  shortest  line-segment  that  can  be  drawn  between  the  lines. 

335.  Theorem.  —  If  each  of  two  intersecting  planes  is  perpen- 
dicular to  a  third  plane,  their  line  of  intersection  is  perpendicu- 
lar to  that  plane, 

R\ B 


Hypothesis.  Plane  P^±  plane  MN\  plane  T^aS'X  plane 
MN ',  and  planes  PQ  and  RS  intersect  in  line  AB,  which 
meets  iHfiV  at  A. 

Conclusion.     Line  AB  1.  plane  MN. 

Suggestions.  Draw  line  AO A.  plane  MN.  Then  AB  ±  plane 
MN  if  AB  and  AO  are  proved  to  coincide.  AB  and  AO  co- 
incide if  J.  (7  is  proved  to  lie  in  P^  and  in  RS.  Hence  begin 
by  proving  the  latter. 

Write  the  proof  in  full. 


310 


SOLID   GEOMETRY 


336.  Theorem.  —  Every  point  in  the  plane  which  bisects  a 
diedral  angle  is  equally  distant  from  the  faces  of  the  diedral 
angle. 


Hypothesis.  Plane  QR  bisects  diedral  angle  M-QP-N  \ 
A  is  any  point  in  QR  ;  line  AB  A.  plane  NP  and  line  AC  1. 
plane  MP. 

Conclusion.   AB  =  AQ. 

Proof.  1.  Plane  QR  bisects  diedral  angle  M-QP-N  \  A  is 
any  point  in  QR  ;  line  AB ±  plane  NP  and  line  AQ  1.  plane 
MP.  Hyp. 

2.  AB  and  J.  (7  determine  a  plane.  §  298 

3.  This  plane  intersects  planes  NP.,  QR.,  and  MP  in  lines 
DB,  DA,  and  BO,  respectively.  §  301 

4.  The  plane  which  is  determined  by  AB  and  ACis  per- 
pendicular to  NP  and  MP.  §  331 

5.  .-.  the  plane  which  is  determined  by  AB  and  AC  is 
perpendicular  to  edge  QP.  §  335 

6.  .-.  DB  ±  QP,  DA  ±  QP,  BC±  QP.  §  302 

7.  .'.  Z  BBA 'dnd  Z.ABO  are   plane    angles    of   diedral 
angles  N-QP-R  and  R-BP-M,  respectively.  §  326 

8.  .-.  ZBDA^ZABO.  §  328 

9.  In  A  ABB  and  A  A  CD,  AD  >  is  common,  and  Z  ABD 
and  Z  ACD  are  right  angles.  §  302 

10.  .'.AABD^AACD.  §68 

11.  .-.  AB  =  AC.  Def.  congruence 


LINES  AND   PLANES   IN    SPACE  311 

337,  Theorem. —  Ani/  point  within  a  diedral  angle^  and 
equally  distant  from  its  faces y  lies  in  the  plane  which  bisects  the 
diedral  angle. 

Suggestion.  If.^  is  a  point  within  diedral  angle  M-QP-N^ 
equally  distant  from  MP  and  iVP,  let  QR  be  the  plane  de- 
termined by  A  and  edge  QP.  Prove  that  plane  QB  bisects 
diedral  angle  M-QP-N.     Write  the  proof  in  full. 

EXERCISES 

1.  Any  point  not  in  the  bisecting  plane  of  a  diedral  angle  is  unequally 
distant  from  its  faces. 

2.  The  locus  of  points  within  a  diedral  angle  and  equidistant  from 
the  faces  is  the  bisecting  plane  of  the  diedral  angle. 

3.  Find  the  locus  of  points  equidistant  from  two  given  points  and 
also  equidistant  from  the  faces  of  a  diedral  angle. 

4.  Find  the  locus  of  points  equidistant  from  two  parallel  planes  and 
also  equidistant  from  the  faces  of  a  diedral  angle. 

5.  Find  the  locus  of  points  at  a  given  distance  from  a  given  plane 
and  also  equidistant  from  the  faces  of  a  given  diedral  angle. 

6.  Find  the  locus  of  points  equidistant  from  the  faces  of  a  given  die- 
dral angle  and  also  equidistant  from  the  faces  of  another  given  diedral 
angle. 

7.  Find  a  point  in  a  given  plane  that  is  equidistant  from  the  faces  of 
a  given  diedral  angle  and  also  at  a  given  distance  from  a  given  plane. 

8.  Find  a  point  in  a  given  plane  that  is  equidistant  from  two  given 
points  and  also  equidistant  from  the  faces  of  a  given  diedral  angle. 

9.  The  bisecting  plane  of  any  diedral  angle,  if  produced  through  the 
edge,  also  bisects  the  vertical  diedral  angle. 

10.  If  two  planes  intersect,  forming  two  adjacent  diedral  angles,  the 
planes  which  bisect  these  adjacent  angles  are  perpendicular  to  each  other. 

338.  Projections.  —  The  projection  of  a  point  upon  a  plane  is 
the  foot  of  the  perpendicular  from  the  point  to  the  plane. 

The  projection  of  a  given  line  upon  a  plane  is  the  line  con- 
taining the  projections  upon  the  plane  of  all  points  of  the 
given  line. 


312  SOLID   GEOMETRY 

339.  Theorem.  —  The  projection  of  a  straight  line  upon  a 
plane  is  the  straight  line  determined  hy  the  projections  of  any 
two  of  its  points  upon  the  plane. 


Hypothesis.  A  and  B  are  any  two  points  of  straight  line 
AB ;  O  and  D  are  the  projections  of  A  and  B^  respectively, 
upon  plane  MN. 

Conclusion.  The  projection  of  AB  upon  MN  is  straight 
line  CD, 

Proof.  1.  C  and  D  are  projections  of  A  and  B^  respec- 
tively, upon  MN.  Hyp- 

2.  .'.  AOl.  plane  ifiV and  BD  ±  plane  MK  §  338 

3.  .'.AQWBB.  §  821 

4.  .*.  ^C'and  BD  determine  a  plane  AD.  §  299 

5.  Planes  AD  and  MN  intersect  in  straight  line  CD.  §  301 

6.  Plane  AD  _L  plane  MN.  §  331 

7.  .•.  OD  contains  the  projections  upon  MN  of  all  points 
of  AB,  §  334 

8.  .*.  the  projection  of  AB  upon  MNis  straight  line  OD. 

§338 

340.  Inclination  of  a  line  to  a  plane.  —  A  straight  line  which 
intersects  a  plane  intersects  its  projection  upon  the  plane. 


Why? 


The    acute  angle   formed   by   a 


A 


straight   line    which    intersects    a  /  — ^^ '^  \ 

plane,  but  is  not  perpendicular  to     ^^-— ^ 

it,  and  its  projection  upon  the  plane  is  called  the  inclination 
of  the  line  to  the  plane,  as  /.  OB  A, 


LINES  AND   PLANES   IN   SPACE  313 


EXERCISES 

1.  Show  that  in  the  special  case  when  a  line  is  perpendicular  to  a 
plane,  its  projection  upon  the  plane  is  a  point. 

2.  If  a  trough  is  placed  directly  beneath  the  edge  of  a  roof,  all  water 
dripping  from  the  edge  of  the  roof  falls  into  it.  Show  how  this  illus- 
trates the  theorem  in  §  339. 

3.  If  a  line-segment  is  parallel  to  a  plane,  its  projection  upon  the 
plane  is  a  line-segment  equal  to  the  given  line-segment. 

4.  A  line-segment  12  in.  long  has  an  inclination  of  30°  with  a  plane. 
Find  the  length  of  its  projection  upon  the  plane.  Find  its  length  if  the 
inclination  is  45°.     If  it  is  60°. 

5.  If  a  straight  line  is  not  perpendicular  to  a  plane,  the  given  line 
and  its  projection  upon  the  plane  determine  a  second  plane  perpendicular 
to  the  first. 

Suggestion.  —  From  a  point  on  the  given  line,  draw  a  line  perpen- 
dicular to  the  given  plane.  Does  the  plane  determined  by  the  given  line 
and  its  projection  contain  this  perpendicular  ? 

6.  The  projections  upon  a  plane  of  two  parallel  lines  which  are  not 
perpendicular  to  the  plane  are  parallel. 

Suggestion.  —  From  a  point  on  each  of  the  given  parallel  lines  draw 
a  line  perpendicular  to  the'  given  plane.  What  relation  have  these  per- 
pendiculars ?     Seek  to  apply  §  317. 

7.  If  two  parallel  lines  intersect  a  plane,  they  are  equally  inclined 
to  it. 

8.  Parallel  line-segments  are  proportional  to  their  projections  upon  a 
plane. 

9.  If  a  straight  line  intersects  two  parallel  planes,  it  is  equally  in- 
clined to  the  two  planes. 

10.  If  two  planes  are  not  perpendicular,  the  projection  upon  one  of 
any  parallelogram  in  the  other  is  a  parallelogram. 

11.  If  two  planes  are  parallel,  the  projection  upon  one  of  any  polygon 
in  the  other  is  a  congruent  polygon. 


314  SOLID   GEOMETRY 

12.  If  a  straight  line  intersects  a  plane,  but  is  not  perpendicular  to  it, 
the  inclination  of  the  line  to  the  plane  is  the  least  angle  made  by  the 
given  line  and  any  line  drawn  in  the  plane  through  its  foot. 

Suggestion.  —  Let  BC  be  the  pro- 
jection of  BA ,  and  C  the  projection  of  ^< 
A,  upon  MN;  and  let  BD  be  any  other 


jection  oi  nJi ,  ana  o  ine  projection  oi  ^^ 

A,  upon  MN;  and  let  BD  be  any  other  y ^       *  \N 

line  in  MN  than  BC  that  passes  through  /         <^^ — J~ — I  \ 

B.  Mark  off  BD  =  BC  and  draw  AD-  ^/        ^  ^^^'  \ 


Prove  Z  CBA  <  Z  DBA . 

13.  If  a  line  intersects  a  plane,  and  is  not  perpendicular  to  it,  the  ob- 
tuse angle  which  it  makes  with  its  projection  upon  the  plane  is  the 
greatest  angle  that  it  makes  with  any  straight  line  drawn  in  the  plane, 
through  the  intersection. 

Suggestion.  —  In  the  figure  of  Ex.  12,  produce  BC  and  BD  through  B. 

341.  Polyedral  angles-  —  The  figure  formed  by  three  or 
more  rays  which  are  drawn  from  the  same  point,  not  more 
than  two  being  in  one  plane,  together  with  the  portions  of 
planes  determined  by  the  pairs  of  adjacent  rays  and  included 
between  them,  is  called  a  polyedral  angle. 

The  rays  are  called  the  edges  of  the  polyedral  angle.  The 
point  from  which  the  rays  are  drawn  is  called 
the  vertex.  The  portions  of  planes  deter- 
mined by  pairs  of  adjacent  rays  and  included 
between  them  are  called  the  faces.  And  the 
angles  formed  in  the  faces  by  the  pairs  of 
adjacent  edges  are  called  the  face  angles  of 
the  polyedral  angle. 

A  polyedral  angle  whose  vertex  is  0  and  whose  edges  are 
OA,  OB,  00,  and  OD  is  named  0-ABOD. 

A  polyedral  angle  is  convex  if  a  plane  which  intersects  all 
of  the  faces  intersects  them  in  line-segments  which  form  a 
convex  polygon. 

A  polyedral  angle  of  three  faces  is  called  a  triedral  angle. 


LINES  AND   PLANES   IN   SPACE  315 

342.    Theorem.  —  Any  face  angle  of  a  triedral  angle  is  less 
than  the  sum  of  the  other  two  face  angles. 


Hypothesis.  Z  A  00  is  the  greatest  face  angle  of  triedral 
angle  0-ABC. 

Conclusion.     Z  AOO <  Z  AOB -\- Z  BOO. 

Proof.  1.  Z  AOO  is  the  greatest  face  angle  of  triedral 
angle  O-ABO.  Hyp 

2.  Draw  OB  in  face  AOO,  making  Z  AOB  =  Z  AOB. 
And  let  a  plane  cut  off  OB  =  OB  and  meet  00  at  0  and  OA 
at  A. 

3.  AO  is  n  common  side  of  A  AOB  and  A  AOB. 

4.  .'.aAOB^aAOB.  §63 

5.  .-.    AB  =  AB.  Def.  congruence 

6.  But  AO  <AB-{-  BO.  §  146 

7.  .-.    BO<BO.  Ax.  VII 

8.  .-.    Z  BOO  <  ZBOO.         '  §  148 

9.  .-.    Z  AOB -\-Z  BOO  or  Z  AOO  <Z  AOB -{-Z  BOO. 

Ax.  VII 

EXERCISES 

1.  Any  face  angle  of  a  polyedral  angle  is  less  than  the  sum  of  all  of 
the  other  face  angles. 

Suggestion.  —  Draw  planes  through  any  one  edge  of  the  angle  and  all 
of  the  other  non-adjacent  edges.     Apply  §  342. 

2.  Any  face  angle  of  a  triedral  angle  is  greater  than  the  difference  be- 
tween the  other  two  face  angles. 


316  SOLID   GEOMETRY 

343.   Theorem.  —  Tfie  sum  of  the  face  angles  of  any  convex 
polyedral  angle  is  less  than  four  right  angles. 


Hypothesis.      0-ABC -"  is  a  convex  polyedral  angle. 

Conclusion.     A  AOB -{- Z.  BOQ+  etc.  <  4  rt.  A. 

Proof.     1.    0-^^(7  •••  is  a  convex  polyedral  angle.      Hyp. 

2.  Let  a  plane  meet  the  edges  at  J.,  B^  C,  etc.  Let  P  be 
any  point  within  polygon  ABO  "-.     Draw  AP^  BP^  OP^  etc. 

3.  Then  Z  OBA  +  Z  OBO  >  Z  PBA  +  Z  PBO, 

ZOCB  +  Z  001)  >  Z  POB  +  Z  POD,  etc. 

§  342 

4.  .-.  by*  adding,  Z  OBA  +  Z  OBO  +  Z  OOB  +  etc.  > 
Z  PBA  +  Z  PBO+  Z  POB  4-  etc.  Ax.  IX 

That  is,  the  sum  of  the  base  angles  of  the  triangles  whose 
common  vertex  is  0  is  greater  than  the  sum  of  the  base 
angles  of  the  triangles  whose  common  vertex  is  P. 

5.  But  the  sum  of  the  angles  of  each  triangle  is  a  st.  Z. 

§48 

6.  .-.  since  the  number  of  triangles  whose  vertex  is  0 
equals  the  number  of  triangles  whose  vertex  is  P,  the  sums 
of  all  angles  of  the  two  sets  are  equal.  Ax.  II 

7.  .-.  the  sum  of  the  angles  whose  vertex  is  0  is  less  than 
the  sum  of  the  angles  whose  vertex  is  P.  Ax.  VIII 

8.  But  the  sum  of  the  angles  at  P  =  4  rt.  A  §  17 

9.  .-.  ZvlO^  +  Z^Oa+etc.  <4rt.A  Ax.  XII 


LINES  AND   PLANES  IN   SPACE  317 

EXERCISES 

1.  The  sum  of  the  face  angles  of  a  triedral  angle  is  320°.  What  is 
the  greatest  value  that  any  one  of  the  face  angles  may  have  ? 

2.  If  the  face  angles  of  a  triedral  angle  are  equal,  what  is  the  greatest 
value  that  each  of  the  face  angles  may  have  ? 

344.   Equal  and  symmetrical  polyedral  angles.  —  Two   polye- 

dral  angles  are  equal  if  they  may  be  made  to  coincide.     Hence 

to  each  face  angle  or  diedral  angle 

in    one    of    two    equal    polyedral 

angles,    there    corresponds   a   like 

angle  equal  to  it  in  the  other,  and 

these  correspondingr  parts  are  ar- 

^  °    ^  Equal  Polyedral  Angles 

ranged  in  the  same  order. 

Two  polyedral  angles  are  called  symmetrical  if  to  each  face 

angle  or  diedral  angle  in  one  there  corresponds  a  like  angle 

equal  to  it  in  the  other,  and  these 

corresponding  parts  are  arranged  in 

reverse  order.   In  general, symmetrical 

polyedral  angles  cannot  be  made  to 

coincide. 

^  .,,,,,  Symmetrical  Polyedral 

Two  symmetrical  polyedral   angles  may  * 

be  compared  to  a  pair  of  gloves.     To  each 

part  of  one  glove  there  corresponds  a  like  part  of  the  other,  but  these 
like  parts  are  arranged  in  reverse  order,  so  that  the  right  hand  cannot 
be  put  into  the  left  glove. 

EXERCISES 

1.  Construct  from  cardboard,  or  stiff  paper,  models  of  two  symmetri- 
cal triedral  angles  whose  face  angles  are  respectively  60°,  45'",  and  30°. 
Cut  as  indicated  in  the  pattern,  fold 

along  dotted   lines,   and   paste.      Can     V     5r~7f^\  /^^   i^ 

they  be  made  to  coincide  ?  \       /'^j.v   vV    /&yj45*^ 

2.  If  the  faces  of  a  polyedral  angle  ^^        \     J>    "C     /        \ 
are  produced  through  the  vertex,  the 
given  angle  and  the  vertical  polyedral  angle  formed  are  symmetrical. 


318 


SOLID   GEOMETRY 


345.  Theorem.  —  If  two  triedral  angles  have  the  face  angles 
of  one  equal  respectively  to  the  face  angles  of  the  other ^  their 
corresponding  diedral  angles  are  equal. 

M 


Hypothesis.      In   triedral    angles   M-ABQ   and   N-DEF, 
Z  AMB  =  Z  BNE,  A  BMO=  A  ENF,  Z.  AMO  =  ZBNF. 
Conclusion.     Angle  B-AM-G  =  angle  E-BN-F,  etc. 
Proof.     1.   Z  AMB  =  Z  BNE,  etc.  Hyp. 

2.  Mark  off  equal  distances  MA,  MB,  MO,  NB,  NE,  NF, 
on  the  edges  of  angle  M~ABO  and  angle  N-BEF.  Draw 
AB,  BO,  AC,  BE,  EF,  BF.  On  AM-dnd  BN,  respectively, 
take  AX=  BY.  Draw  XG  ±  AM  in  face  AMB,  meeting 
AB  Sit  G;  XH  ±AM  in  face  AMC,  meeting  AC  at  R; 
YI±  BN  in  face  BNE,  meeting  BE  at  7;  YJ A.  BNin  face 
BNF,  meeting  BF  at  J.     Draw  (7^  and  IJ. 

3.  Then  A  AMB  ^  A  BNE,  etc.  §  63 

4.  .-.  AB  =  BE,  and  Z  BAM^  Z  EBN,  etc.     Def.  cong. 

5.  .'.AABC^ABEF.   ,  §  76 

6.  .'.  ZBAO=Z  EBF.  '      Def.  cong. 

7.  Also  A  ^(5^X^  A  D/r  and  A  ^^X^  A  i>e7F.     §67 

8.  .-.  Xa  =  YI,  XE=  YJ;  also  Aa  =  BI,AH=  BJ. 

Def.  cong. 

9.  .'.aAGH^ABIJ.  §63 

10.  .-.  aH=  IJ.  Def.  cong. 

11.  .-.  A  aXH  ^  A  lYJ.  §  76 

12.  .-.  Z  aXH:=  Z  lYJ  Def.  cong. 

(To  be  completed  by  the  student.) 


LINES  AND   PLANES   IN   SPACE  319 

346.  Theorem.  —  If  two  triedral  angles  have  the  face  angles 
of  one  equal  respectively  to  the  face  angles  of  the  other^  and 
arranged  in  the  same  order^  the  triedral  angles  are  equal. 

Suggestion.  Use  superposition.  Show  that  by  placing 
the  triedral  angles  so  that  one  pair  of  equal  face  angles  coin- 
cide, the  triedral  angles  may  be  made  to  coincide  through- 
out.    Apply  §  345.     Write  out  the  complete  proof. 

EXERCISES 

1.  If  two  triedral  angles  have  the  face  angles  of  one  equal  respec- 
tively to  the  face  angles  of  the  other,  but  arranged  in  reverse  order,  the 
triedral  angles  are  symmetrical. 

2.  If  two  triedral  angles  have  two  face  angles  and  the  included  diedral 
angle  of  one  equal  respectively  to  two  face  angles  and  the  included  die- 
dral angle  of  the  other,  and  the  equal  parts  arranged  in  the  same  order, 
the  two  triedral  angles  are  equal. 

Suggestion.  —  Superpose  one  triedral  angle  upon  the  other. 

3.  If  two  triedral  angles  have  two  diedral  angles  and  the  included 
face  angle  of  one  equal  respectively  to  two  diedral  angles  and  the  included 
face  of  the  other,  and  arranged  in  the  same  order,  the  two  triedral  angles 
are  equal. 

4.  If  two  triedral  angles  have  two  face  angles  and  the  included  die- 
dral angle  of  one  equal  respectively  to  two  face  angles  and  the  included 
diedral  angle  of  the  other,  but  the  equal  parts  arranged  in  reverse  order, 
the  triedral  angles  are  symmetrical. 

Suggestion.  —  Construct  a  third  triedral  angle  symmetrical  to  one  of 
the  given  triedral  angles.     Compare  the  parts  of  the  three  angles. 

5.  If  two  triedral  angles  have  two  diedral  angles  and  the  included  face 
angle  of  one  equal  respectively  to  two  diedral  angles  and  the  included 
face  angle  of  the  other,  but  the  equal  parts  arranged  in  reverse  order,  the 
triedral  angles  are  symmetrical. 

6.  If  two  face  angles  of  a  triedral  angle  are  equal,  the  diedral  angles 
opposite  them  are  equal. 

7.  If  two  face  angles  of  a  triedral  angle  are  equal,  the  triedral  angle 
is  equal  to  the  symmetrical  triedral  angle. 


320  SOLID   GEOMETRY 

MISCELLANEOUS  EXERCISES 

1.  In  how  many  positions  must  a  spirit  level  be  observed  in  order  to 
determine  if  the  surface  on  which  it  rests  is  horizontal  ?     Why  ? 

2.  A  room  is  10  ft.  high,  16  ft.  wide,  and  20  ft.  long.  Find  the 
length  of  the  shortest  line  that  can  be  drawn  on  the  floor  and  walls  from 
a  lower  corner  to  the  diagonally  opposite  upper  corner. 

3.  Can  a  triedral  angle  have  for  its  faces  a  regular  decagon  and  two 
equilateral   triangles?     Two  regular    octagons  and  a   square?     Why? 

4.  If  the  inclination  of  a  line-segment  to  a  plane  is  45°  and  the  pro- 
jection upon  the  plane  is  16  in.  long,  how  long  is  the  line-segment? 

5.  If  each  of  a  series  of  parallel  planed  intersects  all  faces  of  a  trie- 
dral angle,  the  intersections  form  a  series  of  similar  triangles. 

6.  A  plane  perpendicular  to  each  of  the  faces  of  a  diedral  angle  inter- 
sects them  in  the  sides  of  the  plane  angle  of  the  diedral  angle. 

7.  If  from  any  point  within  a  diedral  angle  perpendicular  lines  are 
drawn  to  the  faces,  the  angle  between  these  lines  is  the  supplement  of 
the  plane  angle  of  the  diedral  angle. 

8.  Find  the  locus  of  points  equidistant  from  three  given  points  that 
are  not  in  one  straight  line. 

9.  Prove  that  a  line  cannot  be  perpendicular  to  each  of  two  inter- 
secting planes. 

10.  In  the  first  figure  of  §  307,  prove  that  OG  Jl_GF  by  means  of 
the  Theorem  of  Pythagoras,  using  the  equations  OG^  +  GE^  —  0E\ 
OE^  +  EF^  =  0F'\  GF^  =  GE^  +  EF\     Add  these  three  equations. 

11.  If  the  edges  of  one  polyedral  angle  are  perpendicular  to  the  faces 
of  another,  the  edges  of  the  second  are  perpendicular  to  the  faces  of  the 
first. 

12.  In  any  triedral  angle  the  three  planes  bisecting  the  three  diedral 
angles  intersect  in  one  line,  all  points  of  which  are  equally  distant  from 
the  three  faces. 

13.  All  points  within  a  triedral  angle  and  equally  distant  from  its 
three  faces  are  in  the  intersection  of  the  bisecting  planes  of  the  diedral 
angles. 

14.  The  locus  of  points  within  a  triedral  angle  and  equally  distant 
from  its  three  faces  is  the  line  of  intersection  of  the  three  planes  which 
bisect  the  diedral  angles  of  the  triedral  angle. 


CHAPTER  XIII 

PRISMS   AND  CYLINDERS 

347  Geometric  solids.  —  A  geometric  solid  is  a  portion  of 
space  which  is  completely  inclosed  or  separated  from  the  rest 
of  space  by  some  kind  of  surface. 

348.  Polyedrons.  —  A  polyedron  is  a  solid  whose  bounding 
surface  consists  of  portions  of  intersecting  planes. 

The  figure  adjoining  is  a 
polyedron.  It  is  bounded  by 
portions  of  six  planes. 

The  portions  of  planes  which 
form  the  bounding  surface  of     .,  '       Polyedron 

a  polyedron  are  called  its  faces. 

The  intersections  of  the  facfes  are  called  its  edges.  The 
intersections  of  its  edges  are  called  its  vertices.  How  many 
edges  are  there  in  the  polyedron  above  ?  How  many  ver- 
tices are  there  ? 

A  line-segment  which  joins  any  two  vertices  of  a  polye- 
dron that  do  not  lie  in  the  same  face  is  called  a  diagonal  of 
the  polyedron. 

If  a  plane  intersects  a  poly- 
edron, the  polygon  formed  by 
the  intersections  of  the  plane 
and  the  faces  is  called  a  sec-  ""  "^  ^^       Section  ^p 

tion  of  the  polyedron.  '  Polyedron 

If  every  section  of  a  poly- 
edron is  a  convex  polygon,  it  is  called  a  convex  polyedron. 
Only  convex  polyedrons  will  be  considered  in  this  volume. 

321 


322 


SOLID   GEOMETRY 


A  polyedron  of  four  faces  is  called  a  tetraedron  ;  one  of  six 
faces,  a  hexaedron  ;  one  of  eight  faces,  an  octaedron  ;  one  of 
twelve  faces,  a  dodecaedron;  one  of  twenty  faces,  an  icosaedron. 
The  following  drawings  show  models  of  these  polyedrons. 


Tbtraedbon       Hexaedron      Octaedron       Dodecaedron      Icosaedron 

EXERCISES 

1?  What  is  the  least  number  of  faces  that  a  polyedron  can  have? 
Edges  ?    Vertices  ?    What  kind  of  polyedron  is  it  ? 

2.  How  many  edges  has  a  hexaedron?  An  octaedron?  A  dodeca- 
edron ?    An  icosaedron  ? 

3.  How  many  vertices  has  a  hexaedron  ?  An  octaedron  ?  A  dodec- 
aedron?    An  icosaedron  ? 

4.  If  E  is  the  number  of  edges,  F  the  number  of  faces,  and  V  the  num- 
ber of  vertices  of  a  polyedron,  showthat  in  each  of  the  five  polyedrons 
named  above  E  +  2  =  F  +  V.  This,  principle  is  known  as  Euler's 
Theorem, 

5.  Show  that  in  a  tetraedron,  if  S  equals  the  sum  of  the  face  angles 
and  V  equals  the  number  of  vertices,  5  =  4(F  —  2)  right  angles. 

Is  this  formula  true  for  a  hexaedron?  For  an  octaedron?  For  a 
dodecaedron  ?    For  an  icosaedron  ? 

349.    Prisms.  —  A  prism  is  a  polyedron  of  which  two  faces 

are  congruent  polygons  in  parallel 

planes,  and  of   which   the  other 

faces  are  parallelograms  each  of 

which  has  sides  of  the  polygons  as 

two  of  its  opposite  sides.  --^mm^^  mm- 

mi      ^  ^       ,  .         -^^.-w^-- -         Right  Prism 

Ine  two  congruent  polygons  in 

parallel  planes  are  called  the  bases,  and  the  other  faces  are 


PRISMS  AND  CYLINDERS  323 

called  the  lateral  faces.     The  intersections  of  the  lateral  faces 
are  called  the  lateral  edges. 

The  perpendicular  distance  between  the  bases  of  a  prism  is 
called  the  altitude  of  the  prism. 

The  sum  of  the  areas  of  the  lateral  faces  of  a  prism  is  called 
the  lateral  area. 

The  section  of  a  prism  which  is  made  by  a  plane  which 
intersects  all  of  the  lateral  edges  and  is  perpendicular  to 
them  is  called  a  right  section 
of  the  prism. 

A  prism  whose  lateral 
edges  are  perpendicular  to 
its   bases  is  called  a  right 


ic:^ 


Drism.  ^^ii^^^to^  Oblique  Prism 

A   prism    whose    lateral 
edges  are  not  perpendicular  to  its  bases  is  called  an  oblique 
prism. 

A  prism  is  called  triangular,  quadrangular,  hexagonal,  etc., 
according  as  its  bases  are  triangles,  quadrilaterals,  hexagons, 
etc. 

350.  Fundamental  properties  of  a  prism.  —  The  following 
important  properties  of  a  prism  are  easily  deduced  from  the 
definitions  above.  The  student  should  draw  figures  and 
reason  out  the  correctness  of  each. 

(1)  The  lateral  edges  of  a  prism  are  parallel  and  equal. 

(2)  The  altitude  of  a  right  prism  is  equal  to  a  lateral  edge 
of  the  prism. 

(3)  The  lateral  faces  of  a  right  prism  are  rectangles. 

(4)  Sections  of  a  prism  made  hy  parallel  planes  cutting  all 
lateral  edges  are  congruent  polygons. 

(5)  Right  sections  of  a  prism  are  congruent  polygons. 


324 


SOLID   GEOMETRY 


EXERCISES 

1.  On  a  piece  of  cardboard,  draw  a  figure 
similar  to  the  adjoining  figure,  making  each 
side  of  each  triangle  3  in.  and  the  rectangles  3 
in.  by  5  in.  Cut  out  the  pattern,  and,  by  fold- 
ing along  the  dotted  lines  and  pasting,  make  a 
model  of  a  triangular  prism. 

2.  What  kind  of  polyedrons  are  the  cells 
which  contain  the  honey  in  the  comb  of  the 
bee?  What  is  the  advantage  of  building  the 
cells  in  this  form  ? 


■\/ 


3.  In  the  form  of  what  kind  of  polyedron  are  lead  pencils  sometimes 
made? 

4.  A  rectangular  room  or  box  is  what  kind  of  prism  ? 

5.  Glass  prisms,  such  as  that  shown 
in  the  drawing,  are  used  in  optical 
instruments  for  changing  the  direction 
of  light  passing  through  them.  Abeam 
of  white  light  passing  through  this 
prism  is  dispersed  or  separated  into  a 
rainbow-colored  band,  gradually  chang- 
ing from  red  at  one  end,  through  orange, 
yellow,  etc.,  to  violet  at  the  other  end. 

Prismatic  pieces  of  glass  are  used 
chandeliers  to  disperse  the  light. 

6.  The  drawing  shows  a  model  of  a 
polyedron  of  which  two  faces  are  con- 
gruent polygons  in  parallel  planes,  and 
of  which  the  other  faces  are  all  parallel- 
ograms.    Why  is  it  not  a  prism  ? 

7.  A  section  of  a  prism  made  by  a  plane  parallel  to  a  base  is  con- 
gruent to  the  base. 

8.  Any  section  of  a  prism  made  by  a  plane  parallel  to  a  lateral  edge 
is  a  parallelogram. 

9.  The  lateral  faces  of  a  right  prism  are  perpendicular  to  the  bases. 


What  kind  of  prism  is  it  ? 

also  as   pendants  or  fringes  on 


PRISMS  AND  CYLINDERS  325 


351.     Truncated    prisms.  — 

The  part  of  a  prism  contained 

between   a   base  and  a  plane 

which  is  not   parallel  to  the 

base    is    called     a    truncated  "^    /^       Truncated 

prism. 


1 


352.  Parallelopipeds.  —  A  prism  whose  bases  are  parallel- 
ograms is  called  a  par- 
allelepiped. 

A     parallelopiped 
whose  lateral  edges  are  PabaiIII^pkd 

perpendicular      to     the 
bases  is  a  right  parallelopiped. 

A  right  parallelopiped  whose  bases  are  rectangles  is  a 
rectangular  parallelopiped. 

A  rectangular  parallelopiped  all  of  whose  faces  are  squares 
is  a  cube. 

353.  Fundamental  properties  of  a  parallelopiped.  —  The  fol- 
lowing properties  of  a  parallelopiped  follow  from  the  defini- 
tions above.  The  student  should  draw  figures  and  reason 
out  the  correctness  of  each. 

(1)  The  opposite  lateral  faces  of  a  parallelopiped  are  con- 
gruent and  parallel. 

(2)  Any  two  opposite  faces  of  a  parallelopiped  may  he  taken 
as  the  bases. 

(3)  Any  right  section  of  a  parallelopiped  is  a  parallelogram, 

EXERCISES 

1.  Find  the  sum  of  all  of  the  face  angles  of  any  parallelopiped. 

2.  The  diagonals  of  a  rectangular  parallelopiped  are  equal. 

3.  How,  by  measuring  the  diagonals  of  a  piece  of  house  framing, 
wsaii  a  carpenter  tell  when  it  is  truly  rectangular? 


326  SOLID   GEOMETRY 

4.  The  square  of  a  diagonal  of  a  rectangular  parallelopiped  is  equal 
to  the  sum  of  the  squares  of  three  concurrent  edges. 

5.  Find  the  diagonal  of  a  cube  whose  edge  is  2  in. 

6.  If  the  edge  of  a  cube  is  e,  find  the  length  of  a  diagonal  of  the 
cube. 

7.  The  diagonal  of  a  face  of  a  cube  is  3  V2.     Find  the  diagonal  of 
the  cube. 

8.  Are  the  diagonals  of  a  cube  perpendicular  to  each  other  ? 

9.  A  suitcase  is  26  in.  long,  15  in.  high,  and  7  in.  thick.     Can  an 
umbrella  which  is  32  in.  long  be  packed  inside  of  it  ? 

10.  Show  that  the  edge,  diagonal  of  a  face,  and  diagonal  of  a  cube 
are  in  the  ratio  of  1 :  \/2  :  \/3. 

11.  The  sum  of  the  squares  of  the  four  diagonals  of  any  parallelopiped 
is  equal  to  the  sum  of  the  squares  of  the  twelve  edges. 

12.  The  diagonals  of  any  parallelopiped  all  meet  at  one  point,  which 
is  the  middle  point  of  each. 

13.  The  intersection  of  the  diagonals  of  any  parallelopiped  and  the 
intersections  of  the  diagonals  of  two  opposite  faces  are  in  a  straight  line. 

354.    Theorem.  —  TTie  lateral  area  of  a  prism  is  equal  to  the 
product  of  a  lateral  edge  and  the  perimeter  of  a  right  section. 


/^ 


Hypothesis.  MNOP  "-  is  a  right  section  of  a  prism,  and 
AF  is  a  lateral  edge. 

Conclusion.     The  lateral  area  =  AF(^M]Sr+NO+  OP+etc). 

Suggestions.  Express  the  area  of  each  lateral  face,  and 
form  the  sum  of  these  area«.     Apply  §  350  (1).     Factor. 


PRISMS  AND  CYLINDERS  327 

355.    Corollary.  —  The  lateral  area  of  a  right  prism  is  equal 
to  the  product  of  the  altitude  and  the  perimeter  of  a  base. 
The  proof  is  left  to  the  student. 

EXERCISES 

1.  Find  the  lateral  area  of  a  rectangular  parallelepiped  whose  length 
is  24  in.,  width  16  in.,  and  height  12  in.     Find  the  total  area. 

2.  Write  the  formula  which  expresses  the  total  area  of  a  rectangular 
parallelepiped  whose  dimensions  are  a,  b,  and  c. 

3.  Find  the  lateral  area  of  a  right  hexagonal  prism  in  which  each 
side  of  a  base  is  8  in.  and  the  altitude  is  20  in. 

4.  Find  the  lateral  edge  of  a  prism  whose  lateral  area  is  714  sq.  in. 
and  perimeter  of  a  right  section  42  in. 

5.  A  gallon  of  paint  costing  $1.50  will  cover  300  sq.  ft.  with  one 
coat.  How  much  will  it  cost  to  paint  both  sides  of  an  8-foot  fence  around 
an  athletic  field  that  is  310  ft.  wide  and  325  ft.  long? 

6.  How  many  square  feet  of  sheathing  are  necessary  to  cover  the 
sides  and  ends  of  a  box  car  34  ft.  long,  8  ft.  wide,  and  7^  ft.  high? 
Add  10  %  for  waste  in  cutting  and  matching. 

7.  In  computing  the  cost  of  lathing  and  plastering  a  room,  the  total 
area  of  wall  space  is  considered,  without  any  deductions  for  openings. 
A  room  is  9^  ft.  high,  15  ft.  wide,  and  20  ft.  long.  Laths  cost  $  6  per 
thousand,  and  are  delivered  in  bundles  of  50  each.  A  bundle  will  lath 
4  sq.  yd.  How  many  bundles  are  required?  (Part  of  a  bundle  cannot 
be  bought.)     What  will  they  cost  ? 

8.  Find  the  lateral  area  of  a  right  triangular  prism,  each  side  of  a 
base  of  which  is  4  in.  and  the  altitude  of  which  is  9  in.  Find  also  the 
total  area. 

9.  Find  the  total  area  of  a  right  triangular  prism  whose  altitude  is 
24  in.  and  whose  base  is  a  right  triangle  in  which  the  sides  forming  the 
right  angle  are  6  in.  and  8  in.,  respectively. 

10.  Find  the  lateral  area  of  a  right  prism  whose  base  is  a  regular 
hexagon  of  which  each  side  is  5  in.  and  the  altitude  is  12  in.  Find 
also  the  total  area. 


328  SOLID   GEOMETRY 

11.  The  lateral  area  of  a  rectangular  parallelepiped  is  1152  sq.  in. 
and  the  total  area  is  1632  sq.  in.  The  height  is  18  in.  Find  the  length 
and  breadth. 

Suggestion.     Use  a  system  of  equations. 

12.  The  total  area  of  a  rectangular  parallelepiped  is  550  sq.  ft.  The 
width  is  twice  as  great  as  the  depth,  and  the  length  is  three  times  as 
great  as  the  depth.     Find  the  depth,  width,  and  length. 

13.  The  total  area  of  a  cube  is  96  sq.  in.     Fii.d  the  edge. 

14.  li  S  represents  the  total  area  of  the  suiface  of  a  cube,  what  repre- 
sents the  length  of  an  edge  ? 

15.  The  lateral  area  of  a  right  prism  whose  base  is  a  regular  hexagon 
is  iidhVd,  where  d  is  the  apothem  of  the  base  and  h  is  the  altitude  of 
the  prism. 

16.  Write  the  formula  which  expresses  the  total  area  of  the  right 
hexagonal  prism  in  Exercise  15. 

17.  The  three  face  angles  at  one  vertex  of  a  parallelepiped  are  each 
60°,  and  the  three  lateral  edges  of  the  triedral  angle  with  that  vertex 
are  2  in.,  4  in.,  6  in.,  respectively.  Find  to  two  decimal  places  the  area 
of  the  entire  surface  of  the  solid. 

356.  Volume  of  a  solid.  —  The  volume  of  a  solid  is  the 
numerical  measure  of  the  solid  (See  §  115),  the  unit  of 
measure  being  a  cube  whose  edge  is  some  linear  unit. 

Thus,  if  a  rectangular  parallelepiped 
contains  a  cube  whose  edge  is  1  unit  ex- 
actly 24  times,  and  the  cube  is  taken  as  the 
unit  of  measure,  the  volume  of  the  paral- 
lelepiped is  24. 

If  an  empty  rectangular  box  is  2  in.  deep,  4  in.  wide,  and  5  in.  long, 
how  many  cubic  blocks  each  1  in.  long  could  be  packed  in  it?  Then 
what  is  the  volume  of  the  box? 

357.  Congruent  and  equal  solids.  —  Two  solids  which  have 
the  same  volume  are  equal.  It  is  evident  that  equal  solids 
need  not  have  the  same  shape. 

Two  solids  which  are  alike  in  all  respects,  so  that  they 
may  be  made  to  coincide,  are  congruent. 


PRISMS  AND  CYLINDERS  329 

358.  Theorem.  —  Two  prisms  are  congruent  if  three  faces 
including  a  triedral  angle  of  one  are  congruent  respectively  to 
three  faces  including  a  triedral  angle  of  the  other^  and  an 
similarly  placed. 

H  P 

}0 


H3rpothesis.  Prisms  AQ-  and  10  have  faces  BE^  BG,  BB 
respectively  congruent  to  faces  JM^  JO^  JL^  and  similarly 
placed. 

Conclusion.     Prism  AG  ^ prism  10. 

Proof.     1.    BE^JM,Ba^JO,BD^JL.  Hyp. 

2.  .-.  Z.FBA=ANJl  ACBF^/.KJN,  Z  CBA  =c 
ZKJL  Def.  congruence 

3.  . •.  triedral  angles  B-A CF  and  J-IKN^ltq  equal.    §  346 

4.  Hence  by  applying  prism  ^6r^  to  prism  /O,  angles 
B-ACF  ^nd  J-IKN may  be  made  to  coincide,  face  BB  coin- 
ciding with  JL,  BE  with  JM,  BG  with  JO,  D  falling  at 
Z,  etc.      .  §  344 

5.  The  lateral  edges  are  parallel.  §  350,  (1) 

6.  .-.  DH^V\\\  fall  along  LP,  §  46 

7.  .-.  planes  (70"  and  KP  coincide.  §  299 

8.  Since  E,  F,  G  coincide  with  M,  iV,  0,  respectively, 
planes  EG  and  MO  coincide,  and  hence  H  and  P  coincide. 

§  297 

9.  Similarly,  the  other  planes  of  corresponding  lateral 
faces  and  the  other  vertices  of  ^JTand  NP  coincide. 

10.  .-.  prism  AG  ^  prism  10.  §  357 


330  SOLID   GEOMETRY 

359.  Corollary  l.  —  Two  right  prisms  having  congruent  bases 
and  equal  altitudes  are  congruent. 

The  proof  is  left  to  the  student. 

360.  Corollary  2.  —  Two  truncated  prisms  are  congruent  if 
three  faces  including  a  triedral  angle  of  one  are  congruent 
respectively  to  three  faces  including  a  triedral  angle  of  the 
other ^  and  are  similarly  placed. 

For  the  steps  in  the  proof  of  §  358  apply  equally  to  two 
truncated  prisms. 

Draw  two  truncated  prisms  satisfying  the  hypothesis,  and 
think  the  steps  of  the  proof. 

EXERCISES 

1.  Two  rectangular  parallelepipeds  are  congruent  if  the  three  edges 
meeting  at  a  vertex  of  one  are  equal  respectively  to  the  three  edges  meet- 
ing at  a  vertex  of  the  other. 

2.  Does  the  plane  determined  by  two  diagonally  opposite  edges  of 
any  parallelopiped  divide  it  into  two  congruent  triangular  prisms? 
Why? 

3.  Does  the  plane  determined  by  two  diagonally  opposite  edges  of  a 
rectangular  parallelopiped  divide  it  into  two  congruent  triangular 
prisms?    Why? 

4.  Two  triangular  prisms  are  congruent  if  the  lateral  faces  of  one  are 
congruent  respectively  to  the  lateral  faces  of  the  other  and  are  similarly 
placed. 

5.  If  a  square  wooden  beam  of  which  the 
ends  are  at  right  angles  to  the  longitudinal 
edges  is  sawed  lengthwise  along  two  diago- 
nally opposite  edges,  the  two  triangular  pieces 
into  which  it  is  cut  are  congruent. 

6.  For  making  a  right  prism  of  any  de- 
sired kind,  it  is  only  necessary  to  have  a  pat- 
tern of  a  base  and  a  lateral  edge.    Why  ? 


PRISMS  AND  CYLINDERS 


331 


361.  Theorem.  —  An,  oblique  prism  is  equal  to  a  right  prism 
having  for  a  base  a  right  section  of  the  oblique  prism  and  for  its 
altitude  a  lateral  edge  of  the  oblique  prism. 


B        C 


Hypothesis.  AI  is  an  oblique  prism  ;  KS  is  a  right  prism, 
having  for  a  base  KLM "-^  a  right  section  of  prism  Al^  and 
its  altitude  KP  equal  to  AF,  a  lateral  edge  of  prism  AL 

Conclusion.      Oblique  prism  AI=  right  prism  ^aS'. 

Suggestions.  It  can  be  proved  that  prism  AI=  prism  KS 
if  it  is  first  proved  that  truncated  prism  ^i^  ^  truncated 
prism  FS.     How  ? 

The  latter  may  be  proved  if  it  is  first  proved  that  faces 
AB^  AL,  AG  are  congruent  respectively  to  faces  FI.,  FQ^ 
FT.     How  ? 

Hence  begin  by  proving  faces  AD,  AL,  AG  congruent 
respectively  to  faces  FI,  FQ,  FT. 

Write  the  proof  in  full. 

EXERCISES 

1.  Show  that  the  theorem  in  §  361  may  be  proved  in  two  ways,  by 
using  truncated  prisms  AN  and  FS  in  two  ways. 

2.  Show  that  an  oblique  prism  may  be  changed  into  an  equal  right 
prism  by  cutting  it  along  a  right  section  into  two  truncated  prisms,  then 
interchanging  the  positions  of  the  latter. 

3.  Make  a  model  of  wood  for  illustrating  the  theorem  in  £  361. 


332 


SOLID   GEOMETRY 


362.  Theorem.  —  A  plane  passed  through  two  diagonally  op- 
posite edges  of  any  parallelopiped  divides  it  into  two  equal 
triangular  prisms. 

H 


Hjrpothesis.  BH  is  any  parallelopiped;  plane  AG-  passes 
through  diagonally  opposite  edges  AE  and  (7(r,  forming 
triangular  prisms  ABQEFG  and  ADOEHG. 

Conclusion.     Prism  AB  OEFG  =  prism  AB  QEHG, 

Proof.  1.  Plane  AG  passes  through  diagonally  opposite 
edges  AE  and  OG  of  parallelopiped  BH^  forming  triangular 
prisms  AB  OEFG  and  AD  QEHG.  Hyp. 

2.    Let  MNOP  be  a  right  section  of  BH, 

8.    Then  MNOP  is  a  parallelogram.  §  353,  (8) 

4.  r.  AMNO^AMPO.  §83 

5.  Prism  AB  OEFG  =  3i,  right  prism  with  A  MJSTO  for 
base  and  AE  for  altitude,  and  prism  ABOEIIG  =  aL  right 
prism  with  AMPO  for  base  and  AE  for  altitude.      §  361 

6.  But  these  right  prisms  are  equal.  §  359 

7.  .-.  prism  ABOEFG  =  prism  ABOEHG.  Ax.  I 


EXERCISES 

1.  A  plane  passed  through  two  diagonally  opposite  edges  of  a  rec- 
tangular parallelopiped  divides  it  into  two  congruent  triangular  prisms. 

2.  In  the  figure  of  §  362,  the  plane  passed  through  the  diagonally 
opposite  edges  AE  and  CG  and  the  plane  passed  through  the  diagonally 
opposite  edges  BF  and  DH  divide  parallelopiped  BH  into  four  equal 
triangular  prisms. 


PRISMS  AND  CYLINDERS 


333 


3.  A  bin  in  the  form  of  a  rectangular  parallelepiped  that  holds  840 
bu.  of  grain  is  divided  into  four  compartments  by  two  vertical  diagonal 
planes.     How  many  bushels  in  each  compartment? 

363.  Volume  of  a  rectangular  parallelepiped.  The  lengths 
of  the  three  edges  of  a 
rectangular  parallelopiped 
which  meet  at  any  vertex 
are  called  the  dimensions  of 
the  parallelopiped. 

If  the  dimensions  of  a 
rectangular  parallelopiped 

are  2,  3,  and  5  linear  units,  respectively,  it  is  evident  that 
the  parallelopiped  may  be  divided  into  2  x  3  x  5,  or  30, 
cubes,  each  having  its  edge  equal  to  the  same  linear  unit. 
Hence  if  one  of  these  cubes  is  taken  as  the  unit  of  measure 
of  the  parallelopiped,  the  volume  of  the  parallelopiped 
equals  30. 

In  general,  if  the  dimensions  of  a  rectangular  parallelo- 
piped are  a,  5,  and  c  linear  units,  respectively,  the  number 
of  unit  cubes,  or  the  volume  of  the  parallelopiped,  is  ahc. 


This  relation  also  may  be  proved  to  be  true  when  the 
three  edges  intersecting  at  any  vertex  do  not  have  a  com- 
mon unit  of  measure,  or  when  the  dimensions  can  be  ex- 
pressed only  approximately,  although  the  proof  is  not 
attempted  here.     Hence : 

The  volume  of  any  rectangular  parallelopiped  is  equal  to  the 
product  of  its  three  dimensions. 


334  SOLID   GEOMETRY 

364.  Corollary  l.  —  The  volume  of  any  rectangular  parallelo- 
piped  is  equal  to  the  product  of  its  altitude  and  the  area  of  its 
base. 

The  proof  is  left  to  the  student. 

365.  Corollary  2.  —  The  volumes  of  two  rectangular  paral- 
lelopipeds  having  equal  bases  are  to  each  other  as  their 
altitudes. 

For,  if  M  is  the  volume,  b  the  area  of  the  base,  and  h  the 
altitude  of  one  parallelopiped,  and  if  N  is  the  volume,  b  the 
area  of  the  base,  and  k  the  altitude  of  the  other,  then  by 
§  364, 

M=hb  Q,nd  N=kb. 

TT  1.     J-   -J'        M     hb        h 

Hence,  by  dividing,  —-  =  —  or  -• 

Jy      kb        k 

366.  Corollary  3.  —  The  volumes  of  two  rectangular  parallel- 
opipeds  having  equal  altitudes  are  to  each  other  as  their  bases. 

The  proof  is  left  to  the  student. 

367.  Corollary  4.  —  The  volume  of  a  cube  is  equal  to  the  cube 
of  its  edge. 

The  proof  is  left  to  the  student. 

EXERCISES 

1.  Find,  in  its  lowest  terms,  without  multiplication,  the  ratio  of  the 
volumes  of  two  rectangular  parallelopipeds  whose  dimensions  are  6  ft., 
8  ft.,  5  ft.,  and  9  ft.,  12  ft.,  10  ft.,  respectively. 

2.  A  rectangular  water  tank  is  8  ft.  3  in.  long  and  5  ft.  wide.  How 
many  cubic  feet  of  water  does  it  contain  when  the  water  is  18  in.  deep? 

3.  One  cubic  foot  of  iron  weighs  450  lb.  If  an  iron  bar  is  2\  in. 
thick,  4  in."  wide,  and  6  ft.  6  in.  long,  find  its  weight. 

4.  A  gallon  contains  231  cu.  in.  How  many  gallons  does  a  tank 
hold  if  it  is  8  ft.  square  and  6  ft.  high? 

5.  In  a  lot  120  ft.  long  and  66  ft.  wide,  a  cellar  is  to  be  dug  for  a 
building.     The  cellar  is  to  be  44  ft.  Jong,  36  ft.  wide,  and  7  ft.  deep. 


PRISMS  AND   CYLINDERS 


335 


pi 


i4-' 


fe: 


The  earth   removed  is  to   be  used  for  "filling"  the  surrounding  yard 
What  will  be  the  depth  of  the  filling? 

6.  In  steel  construction  work,  such  as  the  con- 
struction of  bridges  and  modern  city  buildings,  the 
weight  of  steel  is  computed  and  not  weighed. 

In  a  certain  building  contract  the  specifications 
require  100  16-foot  beams,  a  right  cross  section  of 
which  is  shown  in  the  margin  (8''  =  8  in.).  Find 
the  weights  of  the  beams  and  the  cost  B.t5fa  pound 
when  put  in  place.     Allow  490  lb.  per  cubic  foot. 

7.  Find  the  weight  of  150  12-foot  steel  beams, 
a  right  cross  section  of  which  is  shown  in  the 
margin. 

8.  In  the  construction  of  a  bridge  there  were 
T-shaped  steel  beams  20  ft.  long  and 
U-shaped  steel  beams  16  ft.  long,  the  right 
cross  sections  of  which  are  shown  in  the 
margin.  Compute  the  weight  of  a  beam  of 
each  kind. 

9.  The  total  area  of  a  cube  is  150  sq.  in.     Find  its  volume. 

10.  The  volume  of  a  given  cube  is  v.  Find  the  volume  of  a  cube  whose 
edge  is  twice  as  long  as  the  edge  of  the  given  cube. 

11.  The  edge  of  a  given  cube  is  e.  Find  to  tenths  the  edge  of  a  cube 
containing  twice  the  volume  of  the  given  cube. 

Note.  —  See  §  185.  It  is  discovered  in  Exercise  11  above  that  the  edge  of 
a  cube  of  which  the  volume  shall  be  just  twice  that  of  a  cube  of  given  edge 
cannot  be  computed  exactly.  The  corresponding  problem  of  construction,  to 
construct  with  compasses  and  unmarked  straightedge  alone  the  edge  of  a 
cube  which  shall  contain  twice  the  volume  of  a  given  cube,  is  one  of  the  three 
famous  impossible  problems  of  geometry.  It  dates  from  the  time  of  the 
ancient  Greeks,  and  is  known  as  the  problem  of  the  Duplication  of  the  Cube. 

One  legend  as  to  the  origin  of  the  problem  is  that  the  Athenians,  who  were 
suffering  from  a  pestilence,  consulted  the  oracle  at  Delos  as  to  how  to  stop 
the  plague.  Apollo  replied  through  the  oracle  that  in  order  to  stop  the  pes- 
tilence the  Athenians  must  double  the  size  of  the  altar  of  the  god  in  Athens. 
This  altar  was  in  the  form  of  a  cube.  A  new  altar  was  constructed  with  an 
edge  twice  as  long  as  the  edge  of  the  old  one.  The  pestilence  then  became 
worse.  The  Athenians  gave  the  problem  to  the  mathematicians,  who  in  time 
proved  it  impossible  of  solution. 


336  SOLID   GEOMETRY 

368.    Theorem.  —  The  volume  of  any  parallelopiped  is  equal 
to  the  product  of  its  altitude  and  the  area  of  its  base. 

D 
C^-^ — A     /  --^^ 


Hypothesis.  P  is  any  parallelopiped  of  which  the  altitude 
is  h  and  the  area  of  the  base  is  a. 

Conclusion.     The  volume  of  P  =  ha. 

Proof.  1.  P  is  any  parallelopiped  of  which  the  altitude  is 
h  and  the  area  of  the  base  is  a.  Hyp. 

2.  Produce  the  edge  AB  and  all  edges  parallel  to  AB. 
On  AB  produced,  mark  off  CD  =  AB,  and  through  0  and  I) 
pass  planes  perpendicular  to  AB,  forming  the  right  parallelo- 
piped Q. 

Produce  the  edge  BU  of  Q  and  the  edges  parallel  to  BB. 
On  BB  produced,  mark  off  FGr  —  BE,  and  through  F  and 
G-  pass  planes  perpendicular  to  BE,  forming  the  rectangular 
parallelopiped  R. 

3.  Then  P  =  ^  and  $  =  i2.  §  361 

4.  .-.  P  =  /2.  Ax.  I 

5.  P,  Q,  and  R  have  a  common  altitude  h.  §  323 

6.  Let  the  areas  of  the  bases  of  Q  and  RhQh  and  c,  re- 
spectively. 

7.  Then  a  =  5  and  5  =  <?.  §  215 

8.  .',  a  =  c.  Ax.  I 

9.  The  volume  of  i2  =  he.  §  364 
10.  .-.  the  volume  of  P  =  ha.  Ax.  XII 
Draw  a  different  figure,  lettering  it  differently,  and  write 

the  proof. 


PRISMS  AND  CYLINDERS 


337 


369.    Theorem.  —  The  volume  of  any  triangular  prism  is  equal 
to  the  product  of  its  altitude  and  the  area  of  its  base. 


Hypothesis.  ABODEF  is  any  triangular  prism  of  which 
the  altitude  is  h  and  the  area  of  the  base  is  a. 

Conclusion.     The  volume  of  ABCDEF=  ha. 

Proof.  1.  The  altitude  of  triangular  prism  ABCDEF  is 
h  and  the  area  of  the  base  is  a.  Hyp. 

2.  Construct  the  parallelepiped  BH  upon  the  edges  AB^ 
BO,  and  BE, 

3.  Then  prism  ABOBEF=^^  parallelepiped  BH.      §  362 

4.  The  volume  of  parallelopiped  BH  =  h  x  ABC  a.  §  368 

5.  And  ABCa  =  2a.  §  83 

6.  .-.  the  volume  of  AB  CBEF  =^xhx2a  =  ha.   Ax.  XII 


EXERCISES 

1.  Find  the  volume  of  a  prism  whose  altitude  is  18  in.  and  whose  base 
is  a  right  triangle  with  legs  9  in.  and  12  in.,  respectively. 

2.  Find  the  volume  of  a  prism  whose  altitude  is  10  in.  and  whose  base 
is  an  equilateral  triangle  with  each  side  6  in. 

3.  Find  the  volume  of  a  triangular  prism  if  the  altitude  is  14  in.  and 
the  sides  of  the  base  8  in.,  6  in.,  6  in.,  respectively. 

4.  Prove  that  the  volume  of  a  triangular  prism  is  equal  to  one  half  of 
the  product  of  any  lateral  face  and  the  perpendicular  to  that  face  from 
any  point  in  the  opposite  edge. 


338 


SOLID   GEOMETRY 


370.    Theorem.  —  The  volume  of  any  prism  is  equal  to  the 
product  of  its  altitude  and  the  area  of  its  base. 


Suggestions.  Any  prism  may  be  divided  into  triangular 
prisms  with  the  same  altitude,  and  with  the  sum  of  their 
bases  equal  to  the  base  of  the  given  prism,  by  passing  planes 
through  the  lateral  edges. 

Express  the  volume  of  each  of  these  triangular  prisms, 
then  the  sum  of  the  volumes,  etc. 

Write  the  proof  in  full. 


EXERCISES 

1.  Cut  out  blocks  of 
wood  as  follows  to  show 
that  the  volume  of  any  par- 
allelopiped  is  equal  to  the 
volume  of  a  rectangular  par- 
all  elopiped  with  the  same 
altitude  and  equal  base  : 

Saw  out  a  block  in  the 
form  of  a  parallelopiped  of 
which  no  face  is  a  rectangle, 
and  let  any  edge  be  AB. 
Saw  this  into  two  blocks  P 
and  Q  by  sawing  at  right  angles  to  AB.  Put  Q  in  the  new  position  R. 
Let  CD  be  an  edge  at  right  angles  to  AB  in  the  new  solid.  Now  saw 
this  into  two  blocks  S  and  T  by  sawing  at  right  angles  to  CD.  Put  S  m 
the  new  position  U.  The  resulting  solid  is  a  rectangular  parallelopiped 
equal  to  the  given  one. 


PRISMS  AND  CYLINDERS 


339 


By  means  of  wooden  or  metal  pins,  these  blocks  may  be  made  to  stay 
together  in  any  position  in  which  they  may  be  placed,  and  hence  used 
as  models  in  the  classroom. 

2.  The  volumes  of  prisms  having  equal  bases  and  equal  altitudes  are 
equal.  ' 

3.  The  volumes  of  two  prisms  having  equal  altitudes  are  to  each 
other  as  tbo  areas  of  the  bases. 

4.  The  volumes  of  two  prisms  having  bases  of  equal  areas  are  to  each 
other  as  the  altitudes. 

5.  The  volume  of  any  oblique  prism  equals  the  product  of  a  lateral 
edge  and  the  area  of  a  right  section. 

6.  The  volume  of  a  right  prism  whose  base  is  a  regular  polygon  is  equal 
to  the  product  of  one  half  of  the  apothem  of  its  base  and  its  lateral  area. 

7.  Show  by  a  geometric  drawing  that  (a:  -]-  yy  =  x^-{-3  x^y  +  3  x-if^ + y^. 

8.  Show  by  a  geometric  drawing  that  x^  —  y^={x  —  y)  (^x^  -\-  xy  +  y^). 

9.  The  altitude  of  a  prism  is  15  in.,  and  its  base  is  a  right  triangle 
whose  legs  are  4  in.  and  8  in.,  respectively.     Find  its  volume. 

10.  The  base  of  a  prism  is  an  equilateral  triangle  each  side  of  which 
is  5  ft.,  and  its  altitude  is  9  ft.     Find  its  volume. 

11.  The  base  of  a  prism  is  a  regular  hexagon  each  side  of  which  is 
3  in.,  and  its  altitude  is  18  in.     Find  its  volume. 

12.  A  prism  is  20  in.  high,  and  its  base  is  a  trapezoid  whose  parallel 
sides  are  12  in.  and  7  in.,  respectively,  and  the  distance  between  them 
8  in.     Find  its  volume. 

13.  A  farmer  has  a  corn  crib  12  ft.  wide  and  16  ft.  long.  The  com 
in  it  is  piled  10  ft.  high  along  one  side,  and  slopes  off  to  a  depth  of  6  ft. 
along  the  opposite  side.  Allowing  2  bu.  to  5  cu.  ft.,  find  how  many 
bushels  it  contains. 

14.  Find  the  cubic  contents  of  a  cement  retain- 
ing wall  120  ft.  long,  and  with  a  right  cross  section 
as  shown  in  the  figure.     (2'  =  2  ft.) 

15.  The  engineer's  plan  shows  a  canal  trapezoidal 
in  cross  section,  9  ft.  deep,  12  ft.  wide  at  the  bottom, 
and  walls  sloping  outward  at  an  angle  of  45°.  The 
canal  is  580  ft.  long.  The  contractor  estimates  that 
it  will  cost  him  30  ^  a  cubic  yard  to  make  the  exca-  * 
vation.     Adding  10  %  for  profit,  what  should  be  his  bid  on  the  job  ? 


2' 


^'ii 

yv-vV': 

A:l'\  •'','•'' 

jC  •'''''.'•'','•••' '' 

/^'''tv'A:"'\-:*A 

2' 

/••  1 1,*". *•'•.*•':'•.' 

cr^ 

■-'^••.••.•.•.•.'••/.•'.•.•.^.\' 

mmm 

10' 


340 


SOLID   GEOMETRY 


of  1  ft. 


16.  Find  the  number  of  cubic  yards  of 
filling  required  for  a  railroad  embankment 
510  ft.  long,  if  its  cross  section  is  trapezoidal, 
10  ft.  high,  16  ft.  wide  at  the  bottom,  and 
10  ft.  wide  at  the  top. 

17.  The  water  in    an    irrigation    ditch   flows  at 
a  second.     The  stream   runs   3  ft.   deep,  is   3   ft. 
wide  at  the  bottom  of  the  ditch  and  6  ft.  wide  at 
the  surface.      Find  the   capacity   of  the   ditch  in 
gallons  per  second.    (Allow  7|  gal.  to  1  cu.  ft.) 

18.  Water  runs  over  a  V-shaped  notch  in  a  dam  at  a  speed  of  3  ft. 
per  second.  The  angle  of  the  notch  is  90°,  and  the  water  runs  to  a  depth 
of  18  in.  How  many  gallons  of  water  does  the  notch  deliver  per  second  ? 

19.  The  resistance,  in  pounds,  of  a  masonry  dam  to  sliding  because  of 
the  pressure  of  water  against  it  is  obtained 
by  multiplying  the  volume  of  the  masonry, 
in  cubic  feet,  by  0.75  times  the  density  or 
weight  per  cubic  foot,  of  the  masonry.  A 
masonry  dam  whose  cross  section  is  a 
trapezoid  15  ft.  high,  3  ft.  wide  at  the  top, 
and  10  ft.  wide  at  the  bottom,  has  a  den- 
sity of  115  lb.  to  the  cubic  foot.     The  dam 

is  60  ft.  long.     Find  the  resistance  which  it  offers  to  sliding. 

371.  Cylinders. — A  cylindrical  surface  is  a  curved  surface 
traced  by  a  straight  line  which  moves  parallel  to  a  given 
fixed  straight  line  and  continually  intersects  a  guiding  curve 
which  is  not  in  the  same  plane  as  the  fixed  line. 

My     "^ 


The  moving  line  in  any  particular  one  of  its  positions  is 
called  an  element  of  the  surface. 


PRISMS  AND  CYLINDERS 


341 


Thus,  if  the  line  AB  moves  so  that  it  is  continually  parallel  to  the  line 
il/iV  and  constantly  intersects  the  curve  CAD,  it  traces  the  cylindrical 
surface  ED.  In  any  particular  position,  ^5  is  an  element  of  the  surface 
ED. 

A  cylinder  is  a  solid  whose  bounding  surface  consists  of  a 
portion  of  a  cylindrical  surface,  called  the  lateral  surface,  and 
portions  of  two  parallel  planes,  called  the  bases,  which  inter- 
sect all  elements  of  the  cylindrical  surface. 

The  altitude  of  a  cylinder  is  the  perpendicular  distance  be- 
tween the  bases. 

A  section  of  a  cylin- 
der is  the  figure  formed 
by  the  intersection  of 
the  cylinder  and  a 
plane.  A  right  section 
of  a  cylinder  is  a  sec- 
tion made  by  a  plane  which  intersects  all  of  the  elements  of 
the  cylindrical  surface  and  is  perpendicular  to  them. 

A  right  cylinder  is  a  cylinder  whose  elements  are  perpendic- 
ular to  the  bases. 


Right  Cylinder 


An  oblique  cylin- 
der is  a  cylinder 
whose  elements 
are  not  perpen- 
dicular to  the 
bases. 

A  circular  cylin- 
der is  a  cylinder  whose  bases  are  circles. 

A  right  circular  cylinder  is  a  right  cylinder  whose  bases  are 
circles. 


Oblique  Cylinder 


372.    Fundamental  properties  of  a  cylinder.  —  The   following 
important  properties  of  a  cylinder  are  easUy  deduced  from 


342 


SOLID   GEOMETRY 


the  definitions  given  above,  and  should  be  established  by  the 
student : 

(1)  The  elements  of  a  cylinder  are  parallel. 

(2)  The  elements  of  a  cylinder  are  equal. 

(3)  The  altitude  of  a  right  cylinder  is  equal  to  an  element. 


EXERCISES 

1.  On  a  piece  of  cardboard  draw  a 
figure  similar  to  the  adjoining  figure,  mak- 
ing the  length  of  the  rectangle  3f  times  the 
diameter  of  the  circles.  Cut  out  the  pat- 
tern, leaving  the  circles  attached  to  the 
rectangle,  fold  and  paste,  and  thus  make  a 
model  of  a  right  circular  cylinder. 

Note.  —  If  the  school  has  a  lathe,  turn  out 
wooden  cylinders  on  the  lathe  for  use  in  class. 

2.  A  straight  line  drawn  through  a  point 
in  a  cylindrical  surface  parallel  to  a  given  element  is  itself  an  element. 

3.  Every  section  of  a  cylinder  made  by  a  plane  containing  an  element 
is  a  parallelogram. 


Suggestions.  —  Let  ABDC  be  the  section  containing  the  given  element 
^JSand  intersecting  the  cylindrical  surface  again  in  some  kind  of  line  CD. 
Prove  that  CD  is  an  element.     First,  draw  the  element  through  C. 

4.  Every  section  of  a  right  cylinder  made  by  a  plane  containing  an 
element  is  a  rectangle. 

5.  Every  section  of  a  cylinder  made  by  a  plane  parallel  to  a  given  ele- 
ment is  a  parallelogram. 


PRISMS  AND  CYLINDERS  343 

373.    Theorem.  —  The  bases  of  any  cylinder  are  congruent. 

C 


Hypothesis.     J.J5(7and  DUF  are  the  bases  of  a  cylinder. 

Conclusion.     ABC  ^  DBF. 

Suggestions.  Let  A  and  B  be  any  two  points  in  the  perim- 
eter of  ABC,  and  (7  any  third  point  of  the  perimeter.  Draw 
the  elements  AD,  BE,  CF,  Draw  AC,  BC,  AB,  DF,  EF, 
DE, 

Prove  that  ^ABC^A  BEF. 

Then  show  that  base  ABC  may  be  superposed  upon  base 
BEF  so  that  A  and  B  coincide  with  2>  and  ^,  respectively, 
and  that  when  this  is  done,  C  falls  upon  jP,  and  hence  all 
points  in  the  perimeter  of  ABC  fall  at  the  same  time  upon 
corresponding  points  in  the  perimeter  of  DEF^  etc. 


EXERCISES 

1.  The  sections  of  a  cylinder  made  by  two  parallel  planes  cutting  all 
of  the  elements  are  congruent. 

2.  Every  section  of  a  cylinder  made  by  a  plane  paral- 
lel to  a  base  is  congruent  to  the  base. 

3.  Every  section  of  a  circular  cylinder  made  by  a 
plane  parallel  to  a  base  is  a  circle. 

4.  The  straight  line  joining  the  centers  of  the  bases 
of  a  circular  cylinder  passes  through  the  centers  of  all 
sections  made  by  planes  parallel  to  the  bases. 


344 


SOLID   GEOMETRY 


f^ 


Inscribed  Prism 


374.  Inscribed  and  circumscribed  prisms  and 
cylinders.  —  A  prism  is  said  to  be  inscribed  in  a 
circular  cylinder  when  the  bases  of  the  prism 
are  inscribed  in  the  bases  of  the  cylinder  and 
the  lateral  edges  of  the  prism  are  elements  of 
the  cylinder. 

The  cylinder  is  said  to  be  circumscribed  about 
the  prism. 

A  prism  is  said  to  be  circumscribed  about 
a  circular  cylinder  when  the  bases  of  the 
prism  are  circumscribed  about  the  bases  of 
the  cylinder  and  the  lateral  edges  are  parallel 
to  the  elements  of  the  cylinder. 

The  cylinder  is  said  to  be  inscribed  in  the 
prism. 

It  is  evident  that  each  lateral  face  of  a  cir- 
cumscribed prism  of  a  cylinder  contains  one 
element  and  no  other  line  nor  point  of  the 
cylinder.  Such  a  plane  is  said  to  be  tangent 
to  the  cylinder. 


375.  Limits.  —  The  following  principles  of  limits  may  be 
assumed : 

If  a  prism  whose  bases  are  regular  polygons  is  inscribed  iri^ 
or  circumscribed  about,  a  circular  cylinder,  and  if  the  number 
of  lateral  faces  is  indefinitely  increased : 

(1 )  The  perimeter  of  a  right  section  of  the  prism  approaches 
the  perimeter  of  a  right  section  of  the  cylinder  as  a  limit. 

(2)  The  lateral  area  of  the  prism  approaches  the  lateral  area 
of  the  cylinder  as  a  limit. 

(3)  The  volume  of  the  prism  approaches  the  volume  of  the 
cylinder  as  a  limit. 


Circumscribed 
Prism 


PRISMS  AND   CYLINDERS 


345 


376.  Theorem.  —  The  lateral  area  of  a  circular  cylinder  is 
equal  to  the  product  of  an  element  and  the  perimeter  of  a  right 
section  of  the  cylinder. 


Hyp. 
Let 


Hypothesis.     /S'=  lateral     area     of     a     circular     cylinder, 
P  =  perimeter  of  a  right  section,  and  ^  =  an  element. 
Conclusion.     S  =  UP. 
Proof.  1.  AS'=lat.  area,  P=perim.  rt.  sec, -^=  element. 

2.  Inscribe  a  prism  whose  base  is  a  regular  polygon. 
«  =  lat.  area  and  p  =  perim.  of  right  section  of  prism. 

3.  Then  the  lateral  edge  of  the  prism  =  E. 
.-.  s  =  Ep. 

But  s  =  S  and  p=  P. 
Since  p  =  P,  then  Ep  =  EP. 
.'.  S=EP. 


§374 

§354 

§375 

§  275,  (2) 

§  275,  (1) 

377.  Corollary.  —  The  lateral  area  of  a  right  circular  cylinder 
is  equal  to  the  product  of  the  altitude  and  the  circumference  of 
its  base. 

378.  Similar  cylinders.  —  Since 
a  right  circular  cylinder  may  be 
formed  by  revolving  a  rectangle 
about  a  side  as  axis,  it  is  called  a 
cylinder  of  revolution.  Two  such 
cylinders  formed  by  revolving 
similar  rectangles  about  homol- 
ogous sides  as  axes  are  called  similar  cylinders  of  revolution. 


346 


SOLID   GEOMETRY 


379.  Theorem.  —  The  lateral  or  total  areas  of  two  similar 
cylinders  of  revolution  are  to  each  other  as  the  squares  of  the 
radii  of  their  bases  or  as  the  squares  of  their  altitudes. 


^^^ 


Hypothesis.  Of  two  similar  cylinders  of  revolution,  the 
lateral  areas  are  S  and  «,  total  areas  T  and  f,  altitudes  S  and 
A,  and  radii  of  bases  R  and  r,  respectively. 

Conclusion.     —=  —  =  -—=—-. 
s       t       r^       h^ 

Proof.  1.  Lateral  areas  are  S  and  «,  total  areas  ^and  t, 
altitudes  IT  and  A,  radii  of  bases  R  and  r,  respectively.    Hyp. 


S_27rRR_RR_R^ff 

s        2  irrh        rh        r       h 

§377 

But  f  =:^. 
h       r 

§127 

'S_R  ^R_R^_S\ 

8       r       r       r^       P 

Ax.  XII 

T_2  7rRH+  2  irR^  _  R^ff-h  R)  _ 
t         2  irrh  +  2  irr^          r(h  +  r) 

r 

H+R 

h  +  r 

§  377,  §  281 

^.       H     R    ir+R     R     H 

Since  -  =  — ,        ;       =  _  =  — . 
h       r      h+r        r       h 

§  118,  (8) 

,  T     R     R     R^     H^ 

"  t      r      r       r2       h^' 

Ax.  XII 

PRISMS  AND  CYLINDERS  347 


EXERCISES 


1.  Prove  the  theorem  in  §  376  by  circumscribing  about  the  cylinder 
a  prism  whose  base  is  a  regular  polygon. 

2.  The  total  area  of  a  right  circular  cylinder  is  equal  to  the  sum  of 
the  altitude  and  radius  of  the  base,  multiplied  by  the  perimeter  of  the  base. 

3.  The  lateral  areas  of  the  cylinders  formed  by  revolving  a  rectangle 
about  each  of  two  adjacent  sides  are  equal. 

4.  If  the  total  area  of  the  surface  of  a  right  circular  cylinder  is  T, 
and  the  radius  of  the  base  R,  what  is  the  altitude  ? 

5.  If  the  lateral  area  of  the  surface  of  a  right  circular  cylinder  is  S, 
and  the  altitude  H,  find  the  radius  of  the  base. 

6.  Find  the  lateral  area  and  also  the  total  area  of  a  right  circular 
cylinder  whose  altitude  is  6  in.  and  diameter  of  base  4  in. 

7.  How  many  square  inches  of  tin  are  required  for  making  an  open 
cylindrical  pail  10  in.  in  diameter  and  12  in.  deep  ? 

8.  The  altitudes  of  two  similar  cylinders  of  revolution  are  3  in.  and 
4  in.,  respectively,  and  the  lateral  area  of  the  smaller  is  45  sq.  in.  Find 
the  lateral  area  of  the  larger. 

9.  The  total  areas  of  two  similar  cylinders  of  revolution  are  64  sq. 
in.  and  144  sq.  in.,  respectively,  and  the  diameter  of  the  smaller  is  6  in. 
Find  the  diameter  of  the  larger. 

10.  The  total  area  of  a  right  circular  cylinder  is  192  sq.  in.,  and  the 
altitude  is  8  in.     Find  the  diameter. 

11.  In  punching  round  holes  through  metal  plates,  the  pressure 
exerted  by  the  punch,  in  pounds,  in  the  ordinary  run  of  work,  must  be 
60,000  times  the  area,  in  square  inches,  of  the  cylindrical  surface  of  the 
hole  made.  Find  the  pressure  required  to  punch  a  hole  f  in.  in  diameter 
through  a  steel  plate  |  in.  thick. 

12.  Find  the  pressure  required  to  punch  a  hole  \  in.  in  diameter 
through  a  piece  of  boiler  plate  ^^  in.  thick. 

13.  In  a  steam  engine,  72  flues,  or  cylindrical  pipes,  each  2  in.  in 
outside  diameter  and  14  ft.  long,  convey  the  heat  from  the  fire  box 
through  the  water.  How  much  heating  surface  do  they  apply  to  the 
water  ? 

14.  A  condenser  contains  800  tubes,  each  f  in.  in  diameter  and  6^  ft. 
long.     Find  the  total  area  of  their  cooling  surface. 


348  SOLID   GEOMETRY 

380.  Theorem.  —  The  volume  of  any  circular  cylinder  is 
equal  to  the  product  of  the  altitude  and  the  area  of  the  lase. 

Hypothesis.  V=  volume  of  circular  cylinder,  A  =  area 
of  base,  IT  =  altitude. 

Conclusion.      V  =  HA. 

Suggestions.  Circumscribe  about  the  cylinder  a  prism 
whose  base  is  a  regular  polygon.  Let  v  =  volume  of  prism, 
a  =  area  of  its  base.     Then  proceed  as  in  the  proof  of  §  376. 

Write  the  proof  in  full. 

381.  Corollary.  — If  H  denotes  the  radius  of  the  base  and  H 
the  altitude  of  a  circular  cylinder^  then  V  =  irR^H. 

The  proof  is  left  to  the  student. 

382.  Theorem.  —  The  volumes  of  two  similar  cylinders  of 
revolution  are  to  each  other  as  the  cubes  of  the  radii  of  their 
bases.,  or  as  the  cubes  of  their  altitudes. 

The  proof  is  left  to  the  student.     Proceed  as  in  §  379. 

EXERCISES 

1.  Prove  the  theorem  in  §  380  by  inscribing  a  prism  whose  base  is  a 
regular  polygon. 

2.  The  volume  of  a  right  circular  cylinder  is  equal  to  the  product  of 
the  lateral  area  and  one  half  of  the  radius  of  the  base. 

3.  The  diameter  of  a  well  is  8  ft.,  and  the  water  is  9  ft.  deep.  Allow- 
ing 1\  gallons  to  1  cubic  foot,  find  the  number  of  gallons  of  water  in  the 

4.  An  easy  method  that  may  be  used  for  finding  M  "'H  '|| 
the  volume  of  an  irregular  solid  that  may  be  put  into  -•'*-'"-  '^^><^ 
water  is  to  immerse  it  in  water  in  a  cylindrical  vessel, 
note  the  diameter  of  the  vessel,  and  the  depth  of  the 
water  before  and  after  the  solid  is  put  in,  and  then 
compute  the  volume  of  the  water  displaced  by  the 
solid. 


PRISMS  AND   CYLINDERS 


349 


A  stone  is  immersed  in  a  cylindrical  jar  of  water.  The  diameter 
of  the  jar  is  8  in.,  the  depths  of  the  water  before  and  after  the  stone  is 
put  in  are  5^  in.  and  6|  in.,  respectively.     Find  the  volume  of  the  stone. 

5.  A  jagged  piece  of  iron  is  immersed  in  a  cylindrical  jar  of  water 
whose  diameter  is  10  in.  The  depths  of  the  water  before  and  after  the 
iron  is  put  in  are  7  in.  and  9^  in.,  respectively.  Find  the  volume  of 
the  iron. 

6.  A  farmer  builds  a  silo  18  ft.  in  diameter 
and  32  ft.  high.  If  a  cubic  foot  of  silage  weighs 
25  lb.,  how  many  tons  of  silage  does  the  silo 
hold? 

7.  A  sheet-metal  worker  wishes  to  construct 
a  cylindrical  tank  of  galvanized   iron  that  will 
hold  63^  gal.  and  that  will  fit  into  a  given  space  51  in.  high.     Find  the 
diameter. 

8.  A  wash  boiler  12  in.  deep,  10  in.  wide,  and  20  in.  long,  has  round 
ends,  i.e.  each  end  is  a  half  cylinder.     How  many  gallons  does  it  hold? 

9.  In  practical  work  men  use  many  rules  of  thumb.  It  is  desirable 
often  to  test  the  accuracy  of  these  rules.     For  example : 

To  obtain  the  weight  of  round  iron,  multiply  the  square  of  the  diame- 
ter in  inches  by  the  length  in  feet,  and  by  2.63.  This  gives  the  weight 
in  pounds  approximately.  Test  the  accuracy  of  this  rule  when  applied 
to  a  2-inch  iron  rod  12  ft.  long,  allowing  480  lb.  to  1  cu.  ft. 

10.  It  often  is  found  necessary  in  machine-shop  work  to  compute  the 
weight  of  the  rim  of  a  flywheel. 

The  rim  of  a  flywheel  is  6  in.  wide,  the  outer 
diameter  4  ft.  4  in.,  and  the  inner  diameter  4  ft. 
Find  its  weight.     Allow  480  lb.  to  1  cu.  ft. 

11.  Find  the  weight  of  a  hollow  iron  column 
12  ft.  long,  10  in.  in  outside  diameter,  and  1  in. 
thick,  allowing  480  lb.  to  1  cu.  ft. 

12.  For  irrigating  a  5-acre  field,  water  is  de- 
livered through  an  8-inch  pipe  at  a  speed  of  2  ft. 
a  second.     How  long  will  it  take  to  deliver  2  in.  of  water  over  the  entire 
field? 

13.  In  the  manufacture  of  lard,  a  cylinder  for  cooling  it  is  3^  ft.  in 
diameter  and  8  ft.  long.     As  the  cylinder  revolves,  the  hot  lard  adheres 


350 


SOLID   GEOMETRY 


to  the  surface  to  a  depth  of  |  in.  and  is  taken  off.  How  many  pounds 
of  lard  are  cooled  at  each  revolution?  How  many  in  an  hour,  if  the 
cylinder  makes  5  revolutions  a  minute?     Count  56^  lb.  to  1  cu.  ft. 

14.  A  lard  tank  4  ft.  in  diameter  and  8  ft.  deep  has  a  jacket  around 
it,  on  the  bottom  and  side,  4  in.  from  the  surface.  How  many  gallons 
of  water  will  the  space  between  the  tank  and  jacket  hold? 

15.  A  cylinder  of  a  steam  pump,  used  for  pumping  city  water,  is  2  ft. 
in  diameter  and  3  ft.  long.  It  is  filled  and  emptied  twice  at  each  revo- 
lution of  the  piston.  Find  the  number  of  gallons  delivered  by  the  pump 
in  a  minute  if  the  piston  makes  24  revolutions  a  minute. 

16.  The  figure  represents  a  view  of  a  rain  gauge, 
used  for  measuring  the  amount  of  rainfall.  The  open- 
ing at  the  top  is  12  in.  in  diameter,  and  the  cylindrical 
stem  is  4  in.  in  diameter.  Suppose  that  in  a  rain  the 
stem  is  filled  to  a  depth  of  4  in.  What  is  the  precipi- 
tation, i.e.  what  is  the  depth  of  the  rainfall  on  the  level 
ground  ? 


17.  The  drawing  shows  the  plan  of  a  cross  section  of 
long.  The  inner  diameter  of  the  brickwork  is  8  ft., 
and  the  outer  diameter  9  ft.  6  in.  Allowing  10  % 
for  space  taken  up  by  mortar,  find  how  many 
thousand  bricks  must  be  ordered  for  use  in  its 
construction. 

18.  The  Spaulding  rule  for  computing  the  num- 
ber of  board  feet  in  16-foot  logs  is  as  follows: 


Diameter  in  inches 

12 

16 

20 

24 

28 

Board  feet 

77 

161 

276 

412 

569 

(A  board  foot  is  the  quantity  of  timber  in  a  board  1  ft.  square  and  1  in. 
thick.     It  is  equivalent  to  ^  oi  a  cubic  foot.) 

Find  the  amount,  in  cubic  feet,  that  is  wasted  in  sawing  up  a  20-inch 
log,  if  figured  by  this  rule.     Find  the  per  cent  wasted. 

19.  The  Doyle  rule,  which  is  the  rule  most  generally  used  through- 
out this  country  for  computing  the  number  of  board  feet  that  a  given 
log  will  make  when  sawed,  is  as  follows : 


PRISMS  AND  CYLINDERS 


351 


"Deduct  4  in.  from  the  diameter  of  the  log  as  an  allowance  for  slab; 
square  one  quarter  of  the  remainder;  and  multiply  the  result  by  the 
length  of  the  log  in  feet." 

According  to  this  rule,  find  the  amount,  in  cubic  feet,  that  is  wasted  in 
sawing  a  12-foot  log  that  is  25  in.  in  diameter.     Find  the  per  cent  wasted 

20.  A  log  10  ft.  long  and  4  ft.  in  diameter  is  de- 
cayed at  the  center.  The  diameter  of  the  decayed 
part  is  10  in.  Find  the  number  of  cubic  feet  of  good 
timber. 

Note.  — The  amount  left  or  wasted  that  is  required  in 
Exercises  20,  21,  and  22  does  not  refer  to  the  board  feet 
of  lumber,  but  merely  to  that  part  of  the  volume  of  the 
log  left  or  wasted. 

21.  A  log  3  ft.  in  diameter  and  12  ft.  long  has  a 
defect  on  the  surface,  due  to  sun  scald,  which  causes 
a  waste  of  a  slab.  The  defect  extends  over  one  fourth 
of  the  surface.  Find  the  number  of  cubic  feet  in  the 
slab  wasted. 

Suggestion.  —  The  area  of  a  right  section  is  the  difference  between  a 
sector  of  a  circle  and  a  triangle. 

22.  A  log  4  ft.  in  diameter  and  16  ft.  long  has  a 
defect  due  to  decay  which  causes  a  waste  of  a  part  of 
the  log  whose  right  section  is  a  sector  of  a  circle  of 
which  the  angle  at  the  center  is  60°.  Find  the  num- 
ber of  cubic  feet  in  the  part  wasted. 

23.  The  "  Inscribed  Square "  rule,  a  rule  of 
thumb  used  in  lumbering,  gives  the  cubic  contents 
of  square  pieces  of  timber  that  can  be  cut  from  cylindrical  logs.  The 
width  of  the  square  piece  is  obtained  by  multiplying  the  diameter  of  the 
log  by  17,  and  dividing  the  result  by  24.     Is  this  accurate? 


MISCELLANEOUS   EXERCISES 

1.  A  section  of  a  tetraedron  made  by  a  plane  parallel  to  any  face  is 
a  triangle. 

2.  The  section  of  a  tetraedron  made  by  a  plane  which  is  determined 
by  the  middle  points  of  any  three  edges  that  do  not  all  meet  at  one  vertex 
is  a  parallelogram. 


352 


SOLID   GEOMETRY 


3.  The  line-segments  which  join  the  middle  points  of  opposite  edges 
of  any  tetraedron  intersect  at  one  point. 

4.  Every  pair  of  lateral  edges  of  a  prism  determines  a  plane  which  is 
parallel  to  every  other  lateral  edge  of  the  prism. 

5.  The  upper  base  of  a  truncated  paral- 
lelopiped  is  a  parallelogram. 

6.  The  sum  of  two  opposite  lateral  edges 
of  a  truncated  parallelopiped  is  equal  to  the 
sum  of  the  other  two  lateral  edges. 

Suggestion.  —  Compare  AE -\-  CG  with 
MN,  and  compare  BF  +  DH  with  MN. 

7.  The  perpendicular  drawn  to  the  lower 
base  of  a  truncated  right  triangular  prism  from  the  intersection  of  the 
medians  of  the  upper  base  is  equal  to  one 
third  of  the  sum  of  the  three  lateral  edges. 

Suggestion.  —  Let  M  be  the  middle 
point  of  DP.  Draw  MN±ABC.  Now 
express  PQ  in  terms  of  MN  and  GO',  also 
express  MN  in  terms  of  DA  and  PQ. 
Eliminate  MN. 

8.  A  man  contracted  to  excavate  a  cellar 
at  80/*  per  cubic  yard.  The  location  was 
on  sloping  ground,  so  that  the  depth  of  the 
cellar  at  the  upper  side  was  8  ft.  and  at  the  lower  side  4  ft.  The  length  of 
the  cellar  from  the  front  or  lower  side  to  the  back  or  higher  side  was  32  ft. 
and  the  width  was  28  ft.    What  price  was  the  contractor  paid  for  the  work  ? 

9.  In  the  manufacture  of  many  cylindrical  articles  from  sheet  metal, 
such  as  can  lids,  shoe-polish  boxes,  etc.,  circular  blanks  are  first  cut  from 
flat  sheet   metal.       These   are  then 

pressed  into  the  required  shape  in  a 

die.    In  some  large  factories  one  man 

is  kept  busy  computing  the  sizes  of 

blanks.      The  computation  is  based 

upon  the  assumption  that  the  total  area  of  the  finished  article  equals  the 

area  of  the  circular  blank.       The   modification  necessary,  due  to  the 

stretching  of  the  metal,  is  afterward  found  by  trial. 

The  lid  of  a  3-pound  lard  bucket  is  5\  in.  in  diameter,  and  has  a 
^-inch  flange.     Find  the  diameter  of  the  blank. 


PRISMS  AND   CYLINDERS 


353 


10.  A  small  sample  vaseline  box  is  |  in.  deep  and  1|  in.  in  diameter. 
Find  the  diameter  of  the  blank  from  which  it  is  made. 

11.  A  Shinola  shoe-polish  box  lid  is  2|  in.  in  diameter,  and  has  a 
f-inch  flange.     Find  the  diameter  of  the  blank  from  which  it  is  made. 

12.  A  Rumford  baking-powder  can  lid  is  3.1  in.  in  diameter,  and  has 
a  ^-inch  flange.     Find  the  diameter  of  the  blank  from  which  it  is  made. 

13.  In  the  manufacture  of  door  hinges,  the  blank  shown  in  the  middle 
figure  is  first  cut  from  flat  sheet  metal,  and  then  bent  into  the  form 
shown  at  the  right  by  forcing 

it  into  a  die.  The  die  con- 
sists of  a  heavy  piece  of  metal 
into  which  is  cut  a  vertical 
slot  just  wide  enough  to  admit 
the  blank  edgewise.  This 
vertical  slot  terminates  in  a 
cylindrical  hole.  By  great 
pressure  against  its  edge,  the  blank  is  forced  into  the  slot,  and  follows 
around  the  wall  of  the  hole,  thus  being  bent  into  the  required  form. 

The  size  of  blank  that  must  be  cut  is  first  computed  theoretically,  and 
because  of  the  stretching  on  one  side  of  the  metal  and  compression  on 
the  other,  modifications  to  be  made  in  the  dimensions  are  found  by  trial. 

The  flat  part  of  a  steel  door  hinge  is  to  be  4  in.  long  and  If  in.  wide, 
and  the  outside  diameter  of  the  cylindrical  part  ^  in.  Find  the  size  that 
the  blank  must  be  made. 

14.  In  a  small  brass  hinge  used  in  the  manufacture  of  furniture,  the 
flat  part  is  to  be  1^  in.  by  f  in.,  and  the  outside  diameter  of  the  cylin- 
drical part  ^^  in.     Find  the  dimensions  of  the  blank  that  must  be  cut. 

15.  The  flat  part  of  a  small  steel  hinge  is  1  in.  by  4  in.,  and  the  outside 
diameter  of  the  cylindrical  part  .2  in.    Find  the  dimensions  of  the  blank. 

16.  Metal  tubes  are  made  by  a  process  similar  to  that  used  in  the 
manufacture  of  hinges,  except 
that  the  die  is  slightly  different, 
as  shown  in  the  drawing.  Find 
the  dimensions  of  the  blanks  for 
making  brass  tubes  used  as  cur- 
tain poles  that  are  22  in.  long  and 
f  in.  in  diameter. 

How  many  square  feet  of  sheet  brass  will  it  take  to  make  1000  ? 


354  SOLID   GEOMETRY 

17.  In  the  manufacture  of  brass  chandeliers,  a  tube  used  is  1  in.  iD 
diameter  and  3  ft.  long.  How  much  sheet  brass  is  required  to  make 
1000  of  these  tubes? 

18.  How  high  must  a  tomato  can  that  is  to  hold  a  quart  be  made,  if 
its  diameter  is  4  in.  ?     Allow  231  cu.  in.  to  a  gallon. 

19.  Find  the  length  of  a  wire  ^  in.  in  diameter  that  can  be  drawn 
from  a  cubic  foot  of  brass. 

20.  A  boiler  of  an  engine  4  ft.  in  diameter  and  16  ft.  long  is  traversed 
by  60  pipes,  each  3  in.  in  diameter,  which  convey  the  heat  through  the 
water.     How  many  gallons  of  water  does  the  boiler  hold? 

21.  If  the  length  of  a  tube  is  I,  and  the  outer  and  inner  diameters  are 
D  and  d,  respectively,  prove  that  the  volume  equals  ^ttI^D^  —  d^).  If 
the  thickness  of  the  tube  is  t,  show  that  the  volume  equals  |  '7rlt(D  +  d). 

22.  Find  the  edge  of  a  cube  whose  volume  and  area  of  the  entire  sur- 
face contain  the  same  number  of  units. 


CHAPTER   XIV 


PYRAMIDS   AND   CONES 


Pyramid 


383.  Pyramids.  —  A  pyramid  is  a  polyedron  of  which  one 
face,  called  the  base,  is  a  polygon  of  any  number  of  sides,  and 
of  which  the  other  faces 
are  triangles  having  a 
common  vertex. 

The  triangles  are 
called  the  lateral  faces, 
and  their  common  ver- 
tex   the   vertex    of    the 

pyramid.     The  edges  which  meet  at  the  vertex  of  the  pyra- 
mid are  called  the  lateral  edges. 

The  sum  of  the  areas   of  the  lateral  faces  is  called  the 
lateral  area. 

The  perpendicular  distance  from  the 
vertex  to  the  plane  of  the  base  is  called  the 
altitude  of  the  pyramid. 

A  pyramid  is  called  triangular,  quadrangu- 
lar, hexagonal,  etc.,  according  as  its  base  is  a 
triangle,  quadrilateral,  hexagon,  etc. 

A  regular  pyramid  is  a  pyramid  whose  base 
is  a  regular  polygon  and  whose  vertex  lies  in 
the  perpendicular  to  the 
base  at  its  center. 

384.  A  truncated  pyra- 
mid. —  A  truncated  pyra- 
mid is  the  part  of  a 
pyramid  included  between 
its  base  and  a  plane  cutting  all  of  the  lateral  edges. 

366 


Regular  Pyramid 


Truncated 
Pyramid 


356  SOLID   GEOMETRY 

The  base  of  the  pyramid  and  the  section  of  the  cutting 
plane  are  called  the  bases  of  the  truncated  pyramid. 

385.   A  frustum  of  a  pyramid.  —  A  frustum  of  a  pyramid  is 

a  truncated  pyramid  of  which  the  bases  are 
parallel. 

The  altitude  of  a  frustum  of  a  pyramid  is  the 
perpendicular  distance  between  the  bases. 


386.  Fundamental    properties   of   pyramids.  — 

The  following  important  properties  of  pyramids     Frustum  of 

are  easily  deduced  from  the  definitions  given 

above. 

The  student  should  draw  figures,  and  reason  out  the  cor- 
rectness of  each. 

(1)  The  lateral  edges  of  a  regular  pyramid  are  equal, 

(2)  The  lateral  faces  of  a  regular  pyramid  are  congruent 
isosceles  triangles. 

(3)  The  altitudes  of  the  faces  of  a  regular  pyramid  drawn 
from  the  vertex  of  the  pyramid  are  equal. 

(4)  The  lateral  edges  of  a  frustum  of  a  regular  pyramid  are 
equal. 

(5)  The  lateral  faces  of  a  frustum  of  a  regular  pyramid  are 
congruent  trapezoids. 

(6)  The  altitudes  of  the  faces  of  a  frustum  of  a  regular 
pyramid  are  equal. 

387.  Slant  height.  —  The  altitude  of  a  lateral  face  of  a  regu- 
lar pyramid,  drawn  from  the  vertex  of  the  pyramid,  is  called 
the  slant  height. 

The  altitude  of  a  lateral  face  of  a  frustum  of  a  regular 
pyramid  is  called  the  slant  height  of  the  frustum. 


PYRAMIDS  AND  CONES 


357 


EXERCISES 

1.  Draw  on  cardboard  a  figure  similar 
to  the  adjoining  figurie,  making  each  side 
of  the  hexagon  2  in.  Cut  olit  the  pattern, 
and  by  folding  along  the  dotted  lines  and 
pasting,  make  a  model  of  a  regular  hex- 
agonal pyramid. 

2.  If  E,  F,  G,  and  H  are  the  middle 
points  of  the  edges  AB,  AD,  CD,  and  BC, 
respectively,  of  the  triangular  pyramid 
A-BCD,  then  EFGH  is  a  parallelogram. 

Suggestion.  —  What  is  the  relation  of  EH  a^nd  FG  to  ACi    Could  it 
be  proved  by  using  a  different  pair  of  lines  ? 

3.  Find  the  slant  height  of  a  regular  quad- 
rangular pyramid  whose  altitude  is  4  in.  and  each 
side  of  whose  base  is  6  in. 

Suggestions. — Let  OE  be  the  altitude  of  the 
regular  quadrangular  pyramid  0-ABCD,  and  let 
OF  be  the  slant  height.  Draw  FE.  Find  FE. 
Then  find  OF. 

4.  Find  the  lateral  edge  of  the  pyramid  in 
Ex.3. 

5.  Find  the  slant  height  of  a  regular 
hexagonal  pyramid  whose  altitude  is 
12  in.  and  each  side  of  whose  base  is  4  in. 

6.  Find  the  lateral  edge  of  the  pyra- 
mid in  Ex.  5. 

7.  The  altitude  of  a  frustum  of  a 
regular  quadrangular  pyramid  is  8  in. 
and  the  sides  of  the  bases  are  12  in. 
and  16  in.,  respectively.     Find  the  slant  height. 

Suggestions.  —  Find  MO  and  NP.     Then  find  NK  and  MK.    Then 
find  MN. 

8.  Find  tne  lateral  edge  of  the  frustum  in  Ex.  7. 

9.  Find  the  slant  height  of  a  regular  triangular  pyramid  each  side  of 
whose  base  is  3  in.  and  whose  altitude  is  4  in.  .     -  ...... 


358 


SOLID   GEOMETRY 


Suggestion.  —  The  apothem  of  the  base  is  a  leg  of  a  right  triangle 
whose  acute  angles  are  30°  and  60°  respectively.  What  is  the  relation 
between  the  sides  of  such  a  triangle  ? 

10.  Find  the  lateral  edge  of  the  pyramid  in  Ex.  9. 

11.  Find  the  slant  height  of  a  frustum  of  a  regular  triangular  pyramid 
the  sides  of  whose  bases  are  9  in.  and  5  in.,  respectively,  and  whose  alti- 
tude is  4  in. 

Suggestion.  —  Proceed  as  in  Exercise  7. 

12.  Find  the  lateral  edge  of  the  frustum  in  Ex.  11. 

13.  A  work  basket  is  in  the  form  of  a  frustum  of  a  regular  hexagonal 
pyramid  whose  slant  height  is  4|  in.  and 
the  sides  of  whose  top  and  bottom  bases 
are  5^  in.  and  4  in.,  respectively.  The 
sides  and  bottom  are  covered  inside  and 
outside  with  silk.  If  one  half  extra  is 
allowed  for  fullness  in  shirring  the  silk, 
how  much  silk  is  required  for  the  basket?  If  the  silk  is  27  in.  wide, 
find  to  an  eighth  of  a  yard  how  much  of  a  yard  is  required. 

14.  The  figure  shows  the  plan  of  a  square 
roof  in  the  form  of  a  frustum  of  a  pyramid, 
the  upper  base  being  a  flat  deck.  CD  is  18 
ft.,  AB  i^Q  ft.,  and  the  height  of  the  roof,  or 
altitude  of  the  frustum,  is  8  ft.  Find  the 
lengths  that  the  rafters  A  C  and  AE  must  be 
cut  in  building  it. 

Suggestion.  —  AE  is  the  hypotenuse  of  a 
right  triangle  whose  legs  are  6  ft.  and  8  ft. 
respectively. 

15.  The  figure  shows  the  plan  of  a  roof 
a    hexagonal    tower.      OA,    OB,   etc.,   are 
rafters;    CD,  EF,  etc.,  are  jack  rafters.     If 
slope  of  the  roof  is  45°,  and  the  length  of  J  B  is 
8  ft.,  find  the  length  that  each  rafter  must  be  cut 
in  building  it. 

16.  Prove  that  the  perimeter  of  the  mid-section 
of  a  frustum  of  any  pyramid,  made  by  a  plane 

parallel  to  the  bases,  is  equal  to  one  half  of  the  sum  of  the  perimeters 
of  the  bases. 


PYRAMIDS  AND   CONES 


359 


388.  Theorem.  —  The  lateral  area  of  any  regular  pyramid  u 
equal  to  one  half  of  the  product  of  the  slant  height  and  the 
perimeter  of  the  bas^. 


A  B 

H3rpothesis.  0-ABC  •  •  •  is  a  regular  pyramid  ;  aS'  =  its 
lateral  area  ;  Z  =  its  slant  height ;  p  =  the  perimeter  of.  its 
base. 

Conclusion.     S  =  \lp. 

Proof.  The  proof  is  left  to  the  student.  Write  the  proof 
in  full. 

389.  Theorem.  —  The  lateral  area  of  a  frustum  of  a  regular 
pyramid  is  equal  to  one  half  of  the  product  of  the  slant  height 
and  the  sum  of  the  perimeters  of  the  bases. 


Hypothesis,  ^^is  a  frustum  of  a  regular  pyramid  ;  S  = 
its  lateral  area  ;  Z  =  its  slant  height  ;  p  =  perimeter  of  one 
base  and  q  =  perimeter  of  the  other  base. 

Conclusion.     S  =  ^  IQp  +  q) . 

Proof.  The  proof  is  left  to  the  student.  Write  the  proof 
in  full. 


360 


SOLID   GEOMETRY 


EXERCISES 

1.  The  lateral  area  of  a  regular  pyramid  is  equal  to  the  product  of 
the  slant  height  and  the  perimeter  of  a  mid-section  made  by  a  plane 
parallel  to  the  base. 

2.  The  lateral  area  of  a  pyramid  is  greater  than  the  area  of  the  base. 
Suggestion.  —  Draw  line-segments  from  the  foot  of  the  altitude  to 

all  of  the  vertices  of  the  base,  dividing  the  base  into  triangles  which  may 
be  compared  to  the  corresponding  lateral  faces. 

3.  Prove  the  theorem  in  §  388  as  a  corollary  of  the  theorem  in  §  389, 
by  reducing  one  base  of  the  frustum  to  a  point. 

4.  The  slant  height  of  a  regular  pyramid  is  24  ft.,  and  the  base  is  a  tri- 
angle each  side  of  which  is  10  ft.     Find  the  lateral  area.     The  total  area. 

5.  The  base  of  a  regular  pyramid  is  a  hexagon  each  side  of  which  is 
16  in.,  and  the  slant  height  is  20  in.  Find  the  lateral  area.  The  total 
area. 

6.  A  tent  is  made  of  canvas  stretched 
tightly  over  six  poles,  which  are  tied  together 
at  the  top,  and  has  the  form  of  a  regular  pyra- 
mid. The  distance  between  the  feet  of  each 
two  adjoining  poles  is  4|  ft.,  and  the  slant 
height  of  the  tent  is  14  ft.  How  many  square 
yards  of  canvas  are  there  in  the  cover  ? 

7.  Find  the  lateral  area  of  a  regular  quad- 
rangular pyramid  each  side  of  whose  base  is  6 
ft.,  and  whose  altitude  is  8  ft. 

8.  The  Great  Pyramid  of  Egypt,  when  completed,  was  481  ft.  high, 
and  each  side  of  its  square  base  was  764  ft.  long. 
How  many  acres  were  there  in  its  surface  ? 

9.  Find  the  lateral  area  of  a  regular  hex- 
agonal pyramid  each  side  of  whose  base  is  4  ft., 
and  whose  altitude  is  10  ft. 

10.  Each  side  of  the  base  of  a  regular  trian- 
gular pyramid  is  4  ft.,  and  its  altitude  is  6  ft. 
Find  the  lateral  area. 

Suggestion.  — Let  PAB  be  one  face;  let  0 
be  the  center  of  the  base;  and  let  OC±AB. 
First  compute  OC;  then  PC.    See  §  114. 


PYRAMIDS  AND  CONES 


361 


11.  Find  the  lateral  area  of  a  triangular  pyramid  whose  lateral  faces 
ire  all  equilateral  triangles  and  whose  altitude  is  4  ft. 

Suggestion.  —  The  apothem  of  the  base  equals  one  third  of  the  slant 
height. 

12.  Find  the  lateral  area  of  a  frustum  of  a  regular  triangular  pyramid 
the  sides  of  whose  bases  are  42  in.  and  72  in.,  respectively,  and  whose 
slant  height  is  36  in. 

13.  The  pedestal  of  a  marble  column  is  in  the  form  of  a  frustum  of  a 
pyramid  whose  bases  are  regular  octagons  with  sides  3  ft.  and  2  ft.  8  in. 
respectively,  and  whose  slant  height  is  14  in.  How  much  surface  must 
be  polished? 

14.  A  marble  monument  consists  of  a  frustum  of  a  square  pyramid 
whose  bases  are  2  ft.  and  1  ft.  10  in.  square,  respectively,  and  whose  slant 
height  is  5  ft.,  surmounted  by  a  pyramid  whose  slant  height  is  2  ft. 
What  is  the  amount  of  polished  surface  ? 

15.  Find  the  lateral  area  of  a  frustum  of  a 
regular  quadrangular  pyramid  the  sides  of 
whose  bases  are  22  in.  and  12  in.,  respectively, 
and  whose  altitude  is  16  in. 

Suggestions.  —  Let  A  BCD  be  a  lateral  face ; 
let  P  and  Q  be  the  centers  of  the  bases;  let 
PHA.CD  and  QKjlAB\  and  let  HE±QK. 
First  compute  KE,  then  HK. 

16.  Find  the  lateral  area  of  a  frustum  of  a 
regular  triangular  pyramid  the  sides  of  whose 
bases  are  16  ft.  and  12  ft.,  respectively,  and  whose  altitude  is  12  ft. 

17.  Find  the  lateral  area  of  the  frustum  of 
a  regular  hexagonal  pyramid  the  sides  of  whose 
bases  are  9  ft.  and  5  ft.,  respectively,  and  whose 
altitude  is  4  ft. 

18.  A  church  spire  which  is  in  the  form  of  a 
regular  hexagonal  pyramid  is  covered  with  slate. 
Each  side  of  the  base  is  6  ft.,  and  a  lateral  edge  is 
38  ft.  In  figuring  on  the  contract  for  building  the 
spire,  the  contractor  must  estimate  the  amount  of 
slate  required  to  cover  it.  How  many  squares 
(a  square  is  100  sq.  ft.)  of  slate  are  required? 


"^0/ 


362 


SOLID   GEOMETRY 


390.  Theorem.  —  If  a  pyramid  is  cut  hy  a  plane  parallel  to 
the  base :  (1)  the  lateral  edges  and  the  altitude  are  divided 
'proportionally ;  (2)  the  section  is  a  polygon  similar  to  the  base. 


B  C 

Hypothesis.   In  pyramid  0-AB  (7  •  •  •,  plane  of  section  FCrH- 
parallel  to  the  base  and  meets  altitude  OilSf  at  N. 

^      ,    .  ...     OF     Oa  ON 

Conclusion.     ( 1 )    = =  . . .  = ; 

^  ^    OA      OB  OM' 

(2)  FaH ABQ..: 

Proof.     1.    Plane  FEW  plane  AQ. 

.-.  Fa  II  AB,  anw  bo,  etc.,  and  NH 

OF^qa^    ^  ON 

'  '  OA       OB      '"      DM 

Also  Z  FaH=  ZABO.Z  GRJ^  Z  BOB,  etc.    §  317 
Z  OFa  =  Z  OAB,  Z FaO  =  AABO.  §  26 

.-.  since  Z  0  is  common,  A  FOa  ^  AAOB,       §  128 
Similarly,  A  GOR  ^ABOC. 


2. 
3. 

4. 
5. 
6. 

7. 


Hyp. 

MO.  §  315 

§  123,  Ax.  I 


8. 


10. 


11. 


FG 
AB 

Fa 


oa 

OB 

an 


and 


Off 
BO 


oa 

OB 


AB     BO 


Similarly,  |f=g,  etc. 
.-.  FGH ABO^'^. 


Def.  sim.  A 


Ax.  I 


Def.  sim.  poly. 


PYRAMIDS  AND   CONES 


363 


391.  Corollary  l.  —  The  area  of  a  section  of  a  pyramid  par- 
allel to  the  base  is  to  the  area  of  the  base  as  the  square  of  its 
distance  from  the  vertex  is  to  the  square  of  the  altitude. 

The  proof  is  left  to  the  student. 

392.  Corollary  2.  —  If  tivo  pyramids  have  equal  altitudes  and 
equal  bases,  sections  made  by  planes  parallel  to  the  bases  at 
equal  distances  from  the  vertices  are  equal. 


For, 

OR 
EF' 


CD      ON 


AB      oSP 


and 


FV' 


Why 


Hence ^=-^.   Why? 
AB     EF  ^ 

nencQCL^an.   Why? 


EXERCISES 

1.  What  part  of  the  area  of  the  base  of  a  pyramid  is  the  area  of  a 
section  made  by  a  plane  which  is  parallel  to  the  base  and  bisects  the 
altitude  ? 

2.  The  altitude  of  a  pyramid  is  16  in.  and  its  base  is  a  square  12  in. 
on  a  side.  What  is  the  area  of  a  section  parallel  to  the  base  whose  dis- 
tance from  the  vertex  is  12  in.  ? 

3.  If  two  pyramids  with  equal  altitudes  are  cut  by  planes  parallel  to 
the  bases,  and  at  equal  distances  from  their  vertices,  the  sections  have 
the  same  ratio  as  the  bases. 

4.  A  pyramid  18  ft.  high  has  a  base  containing  225  sq.  ft.  How  far 
from  the  vertex  must  a  plane  be  passed  parallel  to  the  base  in  order  that 
the  section  may  contain  64  sq.  ft.  ? 

5.  At  what  point  of  the  altitude  of  a  pyramid  should  a  plane  be  passed 
parallel  to  the  base  so  that  the  section  shall  equal  one  half  of  the  base  ? 

6.  The  base  of  a  pyramid  is  10  yd.  square.  A  plane  parallel  to  the 
base  and  9  yd.  from  the  vertex  cuts  a  section  whose  area  is  36  sq.  yd. 
Find  the  altitude  of  the  pyramid. 


364 


SOLID   GEOMETRY 


393.    Theorem.  —  Two    triangular   pyramids    having   equal 
altitudes  and  equal  bases  are  equal. 


Hypothesis.     Triangular  pyramids   V-ABC  and    W-LMN 
have  equal  altitudes  and  equal  bases. 
Conclusion.      V-ABC=  W-LMN. 
Proof.     1.    Suppose  that  W-LMN  >  V-ABO. 

2.  Let  each  of  the  two  equal  altitudes  be  divided  into  n 
equal  parts  of  length  h.  Through  the  points  of  division  pass 
planes  parallel  to  the  bases,  cutting  the  pyramids  in  A  BEF^ 
A  GrRI,  etc.,  and  A  OPQ,  A  EST,  etc.,  respectively. 

3.  On  A  BEF,  A  GUI,  etc.,  as  upper  bases,  and  with  AB, 
BG-,  etc.,  as  lateral  edges,  construct  prisms  a,  5,  etc. 

4.  On  A  LMN^  A  OPQ^  etc.,  as  lower  bases,  and  with 
iO,  OR^  etc.,  as  lateral  edges,  construct  prisms  x^  y^  etc. 

5.  A  BBF=  AOPQ.A  aHI=  A  EST,  etc.  §  392 

6.  a  =  ^  X  A  BEF,  y  =  hxA  OPQ,  etc.  §  369 

7.  .•.  a  =  y,  5  =  2,  etc.  Ax.  IV 

8.  But  V-ABO  >a-\-h^-  etc.,  and  W-LMN < 
x-\-y-{-z-{-  etc.  Ax.  X 

9.  .-.  W-LMN-V-ABO<x+y-^z-hetG.-(a-hh-^etc.), 
or  W-LMN—  V-ABO < x,  for  the  pyramids  differ  by  less  than 
the  difference  between  the  sums  of  the  two  sets  of  prisms. 

10.    Now,  by  increasing  n  indefinitely,  and  thus  reducing 


PYRAMIDS  AND  CONES 


365 


h  indefinitely,  x  can  be  made  less  than  any  fixed  or  assigned 
quantity  however  small. 

Hence,  whatever  value  W-LMN—  V-ABC  is  assumed  to 
have,  X  may  be  made  less  than  W-LMN—  V~ABC^  which 
contradicts  step  9. 

11.  .*.  the  supposition  is  false,  and  W-LMN i^  not  greater 
than  V-ABO, 

12.  Similarly,  it  may  be  shown  that  V-ABO  is  not  greater 
than  W-LMN. 

13.  .-.  V-ABC  =  W-LMN 


EXERCISES 

1.  Prove  the  theorem  in  §  393  by  assuming  that  the  volume  of  a 
triangular  pyramid  is  the  limit  of  the  sum  of  the  volumes  of  a  series  of 
inscribed  prisms  of  equal  altitudes,  if  the  number  of  prisms  is  indefinitely 
increased. 

Suggestions.  —  Proceed 
as  in  §  393,  except  to  con- 
struct inscribed  prisms  A, 
B,  etc.,  and  M,  N,  etc.,  with 
the  sections  as  upper  bases 
in  both  pyramids. 

Show  that  A  +  B  +  etc. 
=  Jlf  +  iV  +  etc. 

Then  apply  §  275,  (1).  • 

2.  A  pyramid  with  a  parallelogram  as  base  is  divided  into  two  equal 
pyramids  by  a  plane  through  the  vertex  and  a  diagonal  of  the  base. 

3.  How  may  a  given  triangular  pyramid  be  divided  into  four  equal 
triangular  pyramids  ? 

4.  Prove  the  theorem  of  plane  geometry  that  two  triangles  which 
have  equal  bases  and  equal  altitudes  are  equal,  by  a  proof  similar  to  that 
in  §  393. 

Suggestion.  —  Inscribe 
a  set  of  parallelograms  in 
one  triangle,  and  circum- 
scribe a  set  of  parallelo- 
grams about  the  other. 


366  SOLID   GEOMETRY 

394.  Theorem.  —  The  volume  of  a  triangular  pyramid  is 
equal  to  one  third  of  the  product  of  its  altitude  and  the  area  of 
its  base. 


E^ 

_D 

? 

CI 

A*^-^ 

— — -^c 

B 

Hypothesis.  0-ABO  is  any  triangular  pyramid  of  which 
the  altitude  is  h  and  the  area  of  the  base  is  a. 

Conclusion.      0-ABO  =^  ha. 

Proof.  1.  The  altitude  of  0-ABO  is  h  and  the  area  of 
the  base  is  a.  Hyp. 

2.  Construct  prism  ABOEOD  with  base  J.5(7and  lateral 
edge  OB,  Through  ^0  and  0(7  pass  a  plane.  Then  prism 
ABOEOD  is  composed  of  three  triangular  pyramids,  0-ABO^ 
0-AOE,  0-EDO. 

3.  0-AOE  and  0-EBO  have  the  same  altitude  and 
equal  bases,  AAOE  and  A  EDO,  §  83 

4.  .-.  0-AOE=  0-EBO.  §  393 

5.  0-EBO  is  identical  with  0-EOB. 

6.  0-EOB  and  0-ABO  have  the  same  altitude  and 
equal  bases,  A  EOB  and  A  ABO.  §  349 

7.  .:  0-EOB  =  0-ABO  §393 

8.  .-.  0-AOE=  0-ABO.  Ax.  I 

9.  Now  0-ABO -\-  0-AOE -{-  0-E OB  =  prism  ABOE OB. 

Ax.  X 

10.  .-.  3  times  0-ABO ^  prism  ABOEOB.  Ax.  XII 

11.  .'.  0-ABO  =^pvism  ABOEOB.  Ax.  V 

12.  But  prism  ABOEOB  =  ha.  §  369 

13.  .\  0-ABO  =\  ha.  Ax.  XII 


PYRAMIDS  AND   CONES  367 

395.    Theorem.  —  The  volume  of  any  pyramid  is  equal  to  one 
third  of  the  product  of  its  altitude  and  the  area  of  its  base. 


Hypothesis.  0-ABC  •  •  is  any  pyramid  of  which  the 
altitude  is  h  and  the  area  of  the  base  is  a. 

Conclusion.     0-AB O  -••  =\ha. 

Suggestions.  Upon  what  previous  theorem  may  the  proof 
be  made  to  depend?  How  may  0-ABO  •••  be  divided 
into  triangular  pyramids  having  the  same  altitude  as 
0-ABQ  •'^ 

Write  the  proof  in  full. 

396.  Corollary.  —  The  volume  of  any  pyramid  is  equal  to  one 
third  of  the  volume  of  a  prism  with  an  equal  altitude  and  an 
equal  base. 

The  proof  is  left  to  the  student. 

EXERCISES 

1.  Saw  out  a  triangular  prism  of  wood.  Then  cut  this  prism  into 
three  triangular  pyramids,  and  thus  make  a  model  to  illustrate  §  394. 

2.  Construct  of  cardboard  a  prism  and  a 
pyramid  with  equal  bases  and  equal  altitudes, 
leaving  the  base  of  the  pyramid  and  one  base  of 
the  prism  open.  Fill  the  pyramid  level  full  of  dry 
sand,  and  empty  it  into  the  prism.  Repeat  until 
the  prism  is  full.  Does  this  verify  the  corollary 
in  §396? 


368  SOLID   GEOMETRY 

3.  Two  pyramids  having  equal  altitudes  and  equal  bases  are  equal. 

4.  The  volumes  of  any  two  pyramids  with  equal  altitudes  are  to  each 
other  as  the  areas  of  their  bases. 

5.  The  volumes  of  any  two  pyramids  with  equal  bases  are  to  each 
other  as  their  altitudes. 

6.  The  volume  of  a  regular  pyramid  is  equal  to  the  lateral  area  mul- 
tiplied by  one  third  of  the  perpendicular  distance  from  the  center  of  the 
base  to  any  lateral  face. 

7.  Where  must  a  plane  be  passed  through  a  given  pyramid  parallel 
to  the  base  so  that  the  pyramid  cut  off  shall  equal  one  ninth  of  the  given 
pyramid  ? 

8.  A  triangular  pyramid  can  be  constructed  equal  to  any  given 
pyramid  whose  base  is  any  polygon. 

9.  The  volume  of  a  pyramid  equals  the  product  of  the  altitude  and 
the  area  of  a  section  parallel  to  the  base  how  far  from  the  vertex? 

10.  The  line-segments  joining  the  center  of  a  cube  to  the  four  ver- 
tices of  one  face  are  the  lateral  edges  of  a  regular  quadrangular  pyramid 
whose  volume  is  one  sixth  that  of  the  cube. 

11.  The  altitude  of  a  pyramid  is  6  ft.,  and  its  base  is  a  rectangle  5  ft. 
long  and  4  ft.  wide.     Find  the  volume. 

12.  The  altitude  of  a  pyramid  is  18  in.,  and  its  base  is  a  trapezoid 
whose  parallel  sides  are  6  in.  and  14  in.  respectively,  and  the  distance 
between  them  12  in.     Find  the  volume. 

13.  Find  the  volume  of  a  regular  pyramid  whose  altitude  is  24  ft. 
and  base  a  triangle  each  side  of  which  is  12  ft. 

14.  Find  the  volume  of  a  regular  hexagonal  pyramid  whose  altitude 
is  40  in.  and  each  side  of  whose  base  is  15  in. 

15.  A  farmer  has  a  rick  of  corn  piled  out  of  doors  that  is  12  ft.  wide 
at  the  bottom  and  35  ft.  long.  It  tapers  to  a  point  10  ft.  high  in  the 
middle.  How  many  bushels  does  it  contain?  (Count  2  bu.  for  each 
5  cu.  ft.) 

16.  A  farmer  has  a  corn  crib  12  ft.  wide  and  16  ft.  long.  It  is  filled 
with  corn  to  a  depth  of  6  ft.  around  the  walls,  and  heaped  up  in  the 
center  in  the  form  of  a  pyramid  to  a  total  depth  of  9  ft.  How  many 
bushels  does  it  contain  ? 


PYRAMIDS  AND  CONES 


369 


397.  Theorem.  —  The  volume  of  a  frustum  of  a  pyramid  is 
equal  to  one  third  of  the  product  of  its  altitude  and  the  sum  of 
the  areas  of  the  bases  and  the  mean  proportional  between  them, 

O 


Hypothesis.     B  =  area  of  larger  base,  b  =  area  of  smaller 
base,  and  h  =  altitude  of  frustum  Aff  of  any  pyramid. 
Conclusion.     The  volume  of  Aff=  ^h(B  +  b+  ^M), 
Proof.     1 .    B  =  area  of  larger  base,  b  —  area  of  smaller  base, 
h  =  altitude  of  frustum  AH  oi  any  pyramid.  Hyp. 

2.  Let  0-ABO '"  be  the  pyramid  of  which  ^JjTis  a  frus- 
tum, and  let  its  altitude  OiV  meet  base  FH  d^i  M. 

3.  Volume   of    0-ABC -"  =  \B(h  +  OM),   volume   of 

0-Fan^-.  =ib(iOMy  §395 

4.  .-.volume     of     AR  =  I  B(h -{-  OM}- lb(OM)     or 
ihB+lOM(B-b).  Ax.  Ill 


5.   But 


B      ON' 


0]\f 

VB-^b 


and  hence 


Vb^  on 

■y/b        OM 


§  891,  Ax.  VI 


7. 
8. 

9. 
10. 


ON-  OM      _h 
OM       ^^  OM 


V5_      OM      m        §i^«'<;^) 

.  OMiiVB  -  V5)  =  hVb.  Clearing  fract. 

.  multiplying  both  members  by  V5  +  V6, 

Oiltf(^-5)  =  A(VM+6).  Ax.  IV 

.volume  oiAR^^hB  +  ihX^Bb+b).       Ax.  XII 
.  volume  of  Aff=  ^h(B+b  +  V!55).  Factoring 


370 


SOLID   GEOMETRY 


EXERCISES 

1.  The  frustum  of  a  pyramid  is  equal  to  the  sum  of  three  pyramids 
having  a  common  altitude  equal  to  that  of  the  frustum,  and  for  bases 
the  bases  of  the  frustum  and  a  mean  proportional  between  them,  re- 
spectively. 

2.  Show  that  the  formula  for  the  volume  of  a  pyramid  in  §  395  may 
be  obtained  from  the  formula  for  the  volume  of  a  frustum  of  a  pyramid 
in  §  397  by  making  the  smaller  base  of  the  frustum  zero. 

3.  Show  that  the  formula  for  the  volume  of  a  frustum  of  a  pyramid 
in  §  397  would  reduce  to  the  formula  for  the  volume  of  a  prism  if  the 
bases  were  made  equal. 

4.  Find  the  volume  of  a  frustum  of  a  regular  quadrangular  pyramid 
whose  altitude  is  12  in.  and  the  sides  of  whose  bases  are  16  in.  and  8  in., 
respectively. 

5.  Find  the  volume  of  a  frustum  of  a  triangular  pyramid  whose 
altitude  is  16  ft.  and  the  bases  equilateral  triangles  whose  sides  are  10  ft. 
and  4  ft.,  respectively. 

6.  The  base  of  a  granite  monument  is  a  frustum  of  a  pyramid,  of 
which  the  bases  are  squares  whose  sides  are  6  ft.  and  5  ft.,  respectively, 
and  of  which  the  altitude  is  3|  ft.  Granite  weighs  170  lb.  to  the  cubic 
foot.     Find  the  total  weight  of  the  base. 

7.  One  of  "Cleopatra's  Needles,"  quarried 
in  one  piece,  floated  down  the  Nile,  and  erected 
at  Heliopolis,  Egypt,  about  1500  B.C.,  has  been 
removed  to  Central  Park,  New  York.  It  is  a 
frustum  of  a  quadrangular  pyramid,  64  ft.  high, 
8  ft.  square  at  the  base,  and  5  ft.  square  at  the 
top,  surmounted  by  a  pyramid  7  ft.  high.  It  is 
one  of  the  largest  stones  ever  quarried  in  a  single 
piece.  Counting  170  lb.  to  the  cubic  foot,  com- 
pute its  weight. 

8.  A  factory  chimney  is  to  be  constructed  in 
the  form  of  a  frustum  of  a  pyramid  with  a  square  base.  The  height  is 
to  be  140  ft.,  and  the  sides  of  the  bases  are  to  be  24  ft.  and  12  ft.,  re- 
spectively. The  flue  is  to  be  6  ft.  square  throughout  the  entire  height. 
How  many  thousands  of  bricks,  2  in.  by  4  in.  by  8  in.,  must  be  ordered 
to  build  it,  allowing  10  %  of  the  space  for  mortar? 


PYRAMIDS   AND  CONES  371 

9.  A  grain  elevator  is  in  the  form  of  a  frustum  of  a  pyramid  32  ft. 
high.  The  bases  are  square,  with  sides  13  ft.  and  6  ft.,  respectively. 
Allowing  f  bu.  to  1  cu.  ft.,  find  how  many  bushels  of  corn  it  will  hold. 
10,  A  fruit  grower  sells  cherries  in  boxes  5  in.  square  at  the  top,  4  in. 
square  at  the  bottom,  and  3  in.  deep.  Allowing  67.2  cu.  in.  to  a  quart, 
does  one  of  these  boxes  hold  a  quart  of  cherries  ? 

398.  Cones.  —  A  conical  surface  is  a  surface  traced  by  a 
moving  straight  line  which  constantly  intersects  a  fixed  plane 
curve  and  passes  through  a  fixed  point  not  in  the  plane  of 
the  curve.  The  fixed  point  is  called  the  vertex.  The  moving 
line  in  any  position  is  called  an  element  of  the  surface. 


Thus,  if  line  AB  moves  so  that  it  constantly  intersects  the  fixed  plane 
curve  A  CD  and  constantly  passes  through  the  fixed  point  0,  not  in  the 
plane  of  ACD,  it  traces  a  conical  surface.  0  is  the  vertex,  and  ABj  in 
any  one  position,  is  an  element. 

A  conical  surface  consists  of  two  parts,  which  are  traced 
by  the  two  parts  into  which  the  moving  line  is  divided  at  the 
vertex.     These  two  parts  are  called  the  nappes  of  the  surface. 

A  cone  is  a  solid  bounded  by  a  portion  of  one  nappe  of  a 
conical  surface  and  that  part  of  a  plane  cutting  all  elements 
of  the  conical  surface  which  lies  within  the  surface.  The 
portion  of  the  plane  is  called  the  base,  and  the  conical  surface 
is  called  the  lateral  surface  of  the  cone. 

The  perpendicular  distance  from  the  vertex  to  the  base  of 
a  cone  is  called  its  altitude. 


372  SOLID   GEOMETRY 

A    circular  cone    is 

one  whose  base,  is  in- 
closed by  a  circle. 

The     line-segment 
joining  the  vertex  of 

a  circular  cone  to  the  ^  .       Right  Circular  Cone 

center  of  the  base  is 
called  the  axis  of  the  cone. 

A  right  circular  cone  is  a  circular  cone  whose  axis  is  per- 
pendicular to  the  base. 

A  right  circular  cone  is  called  also  a  cone  of  revolution,  for 
if  a  right  triangle  is  revolved  about  one  of  its  legs  as  an 
axis,  it  will  trace  such  a  cone.  Two  cones  of  revolution 
which  are  traced  by  revolving  similar  right  triangles  about 
homologous  legs  as  axes  are  called  similar  cones. 

399.  Corollary.  —  The  elements  of  a  right  circular  cone  are 
equal. 

The  proof  is  left  to  the  student.     Write  the  proof  in  full. 

400.  Slant  height  of  a  cone.  —  The  length  of  an  element  of  a 
right  circular  cone  is  called  the  slant  height  of  the  cone. 

EXERCISES 

1.  Draw  on  paper  or  pliable  cardboard,  a  sector  of  a  circle,  as  shown 
in  the  drawing.     Make  the  radius  4  in.  and  the 
central  angle  less  than  180°.     Cut  out  the  pattern, 
and  by  folding  and  pasting,  make  a  model  of  the 
lateral  surface  of  a  right  circular  cone. 

2.  The  altitude  of  a  right  circular  cone  is  12  ft. 
and  the  radius  of  the  base  is  5  ft.     Find  the  slant  height. 

3.  The  slant  height  of  a  right  circular  cone  is  40  in.  and  the  radius 
of  the  base  is  18  in.     Find  the  altitude. 

4.  The  slant  heights  of  two  similar  cones  of  revolution  have  the  same 
ratio  as  their  altitudes  or  as  the  radii  of  their  bases. 


PYRAMIDS  AND  CONES  373 

401.   Theorem. — A  section  of  the  lateral  surface  of  any  circu- 
lar cone  made  hy  a  plane  parallel  to  the  base  is  a  circle. 

O 


Hypothesis.      0-ABC  is  a  circular  cone  ;    DUFia  a  section 
of  the  lateral  surface  made  by  a  plane  parallel  to  the  base. 
Conclusion.     BFF  is  a  circle. 
Proof.     1.    Draw  axis  OM,  intersecting  plane  DUF  at  N, 

2.  Let  H  and  S  be  any  two  points  on  DFF.  Draw  ele- 
ments OP  and  OQ  through  H  and  S^  respectively.  Let 
planes  OMP  and  OMQ  intersect  plane  of  base  in  MP  and 
MQ  and  plane  of  DFF  in  WE  and  iViS',  respectively. 

3.  Then  ISTE  li  MP.  §  315 

4.  .'.  ZNEO^Z.MPORndZENO=ZPMO.        §26 

5.  Zi20iV^=ZP0iI[f  identically. 

6.  .'.  AOEJSr^AOPM  §128 

rj  EN      ON  r»  4r     .  1 

^-    •••    PM-^m'  Def.  Sim.  poly. 

8.    Similarly,  1^=^. 
^    QM     OM 

g      .    EN ^SN 

'  '  PM     QM 

10.    But  PM=  QM.  §  151,  (2) 

n.    .'.EN=SN  Clearing  fract. 

12.    .  •.  all  points  on  DFF  are  equidistant  from  N,  and 

hence  DEF  is  a  circle.  Def.  O 


Ax.  I 


374  SOLID   GEOMETRY 

402.  Corollary  l.  —  The  axis  of  a  circular  cone  passes  through 
the  center  of  every  section  parallel  to  the  base. 

The  proof  is  left  to  the  student. 

403.  Corollary  2.  —  The  parts  of  the  elements  of  a  right  circu- 
lar cone  included  between  the  base  and  a  plane  parallel  to  the 
base  are  equal. 

The  proof  is  left  to  the  student. 

404.  Conic  sections. — A  section  of  the  lateral  surface  of  a 
right  circular  cone  by  a  plane  is  called  a  conic  section. 

There  are  five  possible  forms  of  conic  sections,  as  follows : 


If  the  cutting  plane  is  parallel  to  the  base,  the  section 
was  proved  in  §  401  to  be  a  circle^  as  shown  in  figure  (1). 

If  the  cutting  plane  intersects  all  elements  but  is  not  paral- 
lel to  the  base,  the  section  is  an  oval  called  an  ellipse^  as 
shown  in  figure  (2).  The  ellipse  is  of  great  interest  and 
importance.  The  orbits  of  the  earth  and  all  other  planets 
are  ellipses. 

If  the  cutting  plane  is  parallel  to  an  element,  the  section 
is  Si  parabola,  as  shown  in  figure  (3).  The  path  of  a  projec- 
tile, such  as  a  stone  thrown  upward,  or  a  bullet  from  a  gun, 
is  a  parabola. 

If  the  cutting  plane  is  parallel  to  the  axis,  the  section  is 
an  hyperbola,  as  shown  in  figure  (4). 

If  the  cutting  plane  passes  through  the  vertex,  the  section 
is  two  straight  lines,  as  shown  in  figure  (5). 


PYRAMIDS  AND  CONES  375 

405.  Frnstuin  of  a  cone.  —  A  frustnm  of  a  cone  is  the  portion 
of  the  cone  lying  between  the  base  and  a  plane  parallel  to 
the  base. 

The  base  of  the  cone  and  the  section  made  by  the  plane 
are  called  the  bases  of 
the  frustum. 

The  perpendicular 
distance  between  the 
bases  is  called  the  alti- 
tude of  the  frustum.  .  J^     Frustum  of  Cone 

The    length   of    an        "     -^~^=  - 
element  included  between  the  bases  is  called  the  slant  heig^ht 
of  the  frustum. 

EXERCISES 

1.  Every  section  of  a  cone  made  by  a  plane  passing  through  the 
vertex  is  a  triangle. 

Suggestion.  —  Prove  that  the  intersections  of  the  plane  and  the  conical 
surface  are  elements. 

2.  Every  section  of  a  frustum  of  a  cone  made  by  a  plane  passing 
through  an  element  is  a  trapezoid. 

3.  Find  the  locus  of  the  centers  of  all  circular  sections  made  by 
planes  parallel  to  the  base  of  a  circular  cone. 

4.  The  radii  of  the  bases  of  a  frustum  of  any  circular  cone  have  the 
same  ratio  as  the  distances  of  the  bases  from  the  vertex  of  the  cone. 

5.  Two  sections  of  a  circular  cone  made  by  planes  parallel  to  the 
base  are  to  each  other  as  the  squares  of  their  dis- 
tances from  the  vertex.  ^ — s^"    "^"7C>v^ 

6.  On  paper  or  pliable  cardboard,  draw  a  figure  \  ^ — ^  J 
similar  to  the  adjoining  figure.     By  cutting  out  the       >v  y 

pattern,  folding  and  pasting,  make  a  model  of  the  ""^ 

conical  surface  of  a  frustum  of  a  right  circular  cone. 

7.  The  altitude  of  a  frustum  of  a  right  circular  cone  is  8  in.,  and  the 
radii  of  the  bases  are  4  in.  and  6  in.,  respectively.  Find  the  slant 
height. 


376 


SOLID   GEOMETRY 


8.  The  altitude,  slant  height,  and  radius  of  the  larger  base  of  a  frus- 
tum of  a  right  circular  cone  are  6  in.,  8  in.,  and  10  in.,  respectively. 
Find  the  radius  of  the  smaller  base. 

9.  A  plane  determined  by  an  element  of 
a  circular  cone  and  a  tangent  to  the  base  con- 
tains no  point  of  the  surface  without  that  ele- 
ment. Such  a  plane  is  said  to  be  tangent  to 
the  cone. 

Suggestion,  —  Let  P  be  any  point  of  the 
plane  without  the  element.  Draw  a  plane 
through  P  parallel  to  the   base.     Prove   that 

this  plane  intersects  the  given  plane  in  a  tangent  to  the  section  of  the 
cone. 

406.  Inscribed  and  circumscribed  pyramids  and  cones.  —  A 
pyramid  is  inscribed  in  a  circular  cone  when  the  base  of  the 
pyramid  is  inscribed  in  the  base  of  the  cone  and  the  two 
have  the  same  vertex. 


/ 

\ 

/  /c-~\  .\ 

^-^ 

Inscribed  Pyramid 


Circumscribed  Pyramid 


A  pyramid  is  circumscribed  about  a  circular  cone  when  the 
base  of  the  pyramid  is  circumscribed  about  the  base  of  the 
cone  and  the  two  have  the  same  vertex.  It  is  evident  that 
when  a  regular  pyramid  is  circumscribed  about  a  right  cir- 
cular cone  the  slant  height  of  the  pyramid  is  equal  to  the 
slant  height  of  the  cone. 

A  frustum  of  a  pyramid  is  inscribed  in  a  frustum  of  a  cir- 
cular cone  when  its  bases  are  inscribed  in  the  bases  of  the 
frustum  of  a  cone. 


PYRAMIDS  AND  CONES  377 


Inscribed  Frustum  Circumscribed  Frustum 

OF  Pyramid  op  Pyramid 

A  frustum  of  a  pyramid  is  circumscribed  about  a  frustum  of  a 
circular  cone  when  its  bases  are  circumscribed  about  the  bases 
of  a  frustum  of  a  cone.  It  is  evident  that  when  a  frustum  of 
a  regular  pyramid  is  circumscribed  about  a  frustum  of  a  right 
circular  cone  the  slant  height  of  the  frustum  of  the  pyramid 
is  equal  to  the  slant  height  of  the  frustum  of  the  cone. 

407.  Limits.  —  The  following  principles  of  limits  are  as- 
sumed: 

(1)  The  lateral  area  of  a  right  circular  cone  (^frustum  of  a 
right  circular  cone)  is  the  limit  approached  by  the  lateral  area 
of  an  inscribed  or  circumscribed  regular  pyramid  (^frustum  of 
a  regular  pyramid)  as  the  number  of  faces  is  indefinitely  in- 
creased. 

(2)  The  volume  of  a  circular  cone  (frustum  of  a  circular  cone) 
is  the  limit  approached  by  the  volume  of  an  inscribed  or  cir- 
cumscribed pyramid  (frustum  of  a  pyramid)  whose  base  is  a 
regular  polygon  as  the  number  effaces  is  indefinitely  increased, 

408.  Principles  of  limits.  —  The  following  general  principles 
of  limits  are  assumed,  as  were  similar  principles  in  §  275  : 

(1)  The  limit  of  the  sum  of  two  or  more  variables  is  the  sum 
of  their  limits. 

(2)  ITie  limit  of  the  product  of  two  or  more  variables  i»  the 
product  of  their  limits. 

(3)  The  limit  of  any  principal  root  of  a  variable  is  that  root 
of  its  limit. 


378  SOLID   GEOMETRY 

409.  Theorem.  —  The  lateral  area  of  a  right  circular  cone  is 
equal  to  one  half  of  the  product  of  its  slant  height  and  the  cir- 
cumference of  its  base. 


Hypothesis.  S  =  the  lateral  area,  O  =  the  circumference 
of  the  base,  and  L  =  the  slant  height  of  a  right  circular 
cone. 

Conclusion.     S  =  -^  LQ, 

Proof.  1.  /S'==the  lateral  area,  (7=  the  circumference 
of  the  base,  and  L  =  the  slant  height  of  a  right  circular 
cone.  Hyp. 

2.  Let  s  =  the  lateral  area  of  a  circumscribed  regular 
pyramid,  and  let  p  =  the  perimeter  of  its  base.  Then  L  = 
the  slant  height  of  the  pyramid  also.  §  406 

3.  .-.  s  =  iLp.  §388 

4.  Now,  if  the  number  of  lateral  faces  of  the  pyramid  is 
increased  indefinitely,  s  =  S  and  p  =  0.  §  407,  §  274 

5.  .-.  lLp  =  ^La  §275,(2) 

6.  r.S=\LO.  §275,(1) 
Using  different  letters,  write  out  the  proof  without  refer- 
ring to  the  book. 

410.  Corollary.  —  If  S  is  the  lateral  area  of  a  right  circular 
cone,  L  the  slant  height,  and  R  the  radius  of  the  base,  then 

The  proof  is  left  to  the  student. 


PYRAMIDS  AND  CONES  379 

EXERCISES 

1.  Can  the  theorem  in  §  409  be  proved  by  inscribing  a  regular 
pyramid  in  the  cone  ? 

2.  Compare  the  lateral  areas  of  a  right  circular  cylinder  and  a  right 
circular  cone  with  equal  bases  and  equal  altitudes  when  the  slant  height 
of  the  cone  is  equal  to  the  diameter  of  the  base. 

3.  Write  the  formula  for  the  total  area  of  a  right  circular  cone. 

4.  Since  the  lateral  surface  of  a  cone  of  revolution  may  be  produced 
by  rolling  up  a  sector  of  a  circle,  prove  the  theorem  in  §  409  by  use  of 
§284. 

5.  The  slant  height  of  a  right  circular  cone  is  8  ft.  and  the  radius  of 
the  base  6  ft.     Find  the  lateral  area.     The  total  area. 

6.  The  altitude  of  a  right  circular  cone  is  12  in.,  and  the  diameter  of 
the  base  14  in.     Find  the  slant  height.     The  lateral  area.     The  total  area. 

7.  The  altitude  oi  a  right  circular  cone  is  18  ft.,  and  the  slant 
height  24  ft.  Find  the  radius  of  the  base.  The  lateral  area.  The  total 
area. 

8.  How  many  square  yards  of  canvas  will  be  required  to  make  a 
conical  tent  whose  altitude  is  12  ft.,  and  diameter  of  base  12  ft.  ? 

9.  A  conical  slate  roof  of  a  tower  is  20  ft.  high,  and  the  diameter 
of  its  base  is  18  ft.     Find  the  area  of  the  slate. 

10.  Find  the  area  of  the  surface  generated  by  revolving  a  right  tri- 
angle about  its  hypotenuse  as  axis,  the  legs  of  the  triangle  being  3  in. 
and  4  in.,  respectively. 

11.  If  a  cone  of  revolution  is  formed  by  revolving  an  equilateral  tri- 
angle about  one  of  the  altitudes  as  an  axis,  the  lateral  area  equals  twice 
the  area  of  the  base. 

Suggestion.  —  Express  the  lateral  area  of  the  cone  in  terms  of  a  side 
of  the  triangle. 

12.  An  equilateral  triangle  each  of  whose  sides  is  6  in.  is  revolved 
about  a  straight  line  through  a  vertex  and  parallel  to  the  opposite  side 
as  axis.  Find  the  area  of  the  whole  surface  traced  by  the  three  sides  of 
the  triangle. 

13.  A  silo  which  is  18  ft.  in  diameter  has  a  conical  roof.  The  rafters 
are  at  an  angle  of  30°  with  horizontal,  and  project  1  ft.  over  the  eaves. 
Find  the  area  of  the  roof. 


380  SOLID   GEOMETRY 

411.  Theorem.  —  The  lateral  area  of  a  frusfMm  of  a  right 
circular  cone  is  equal  to  one  half  of  the  product  of  its  slant 
height  and  the  sum  of  the  circumferences  of  its  bases. 


Hypothesis.  S  =  the  lateral  area,  0  =  circumference  of 
larger  base,  c  =  circumference  of  smaller  base,  and  L  =  slant 
height  of  a  frustum  of  a  right  circular  cone. 

Conclusion.     S  =  ^L(^0 -\-  e^. 

Proof.  1.  S  =  lateral  area,  0  and  c  respectively  =  circum- 
ferences of  bases,  and  L  =  slant  height  of  frustum  of  right 
circular  cone.  Hyp. 

2.  Let  s  =  lateral  area  of  a  circumscribed  frustum  of  a 
regular  pyramid,  and  let  P  and  p  respectively  =  perimeters 
of  its  bases.     Then  L  =  its  slant  height  also.  §  406 

3.  .-.  8  =  -lL(iF  +  p).  §389 

4.  Now,  if  the  number  of  lateral  faces  is  increased  indefi- 
nitely, s  =  S,  F  =  C,  and  p  =  c.  §  407 

5.  .-.  (P+i>)  =  ((7+0.  §408,(1) 

6.  .-.  iX(P+j9)  =  J-Z((7+0-  §275,(2) 

7.  .',S  =  ^LiO+cy  §275,(1) 

412.  Corollary.  If  S  is  the  lateral  area  of  a  frustum  of  a 
right  circular  cone,  L  the  slant  height,  and  B  and  r  respectively 
the  radii  of  the  bases,  then 

S  ^  irLQR  +  r^. 

The  proof  is  left  to  the  student. 


PYRAMIDS  AND  CONES  381 

EXERCISES 

1.  The  lateral  area  of  a  frustum  of  a  right  circular  cone  is  equal  to 
the  perimeter  of  a  mid-section  made  by  a  plane  parallel  to  the  bases, 
multiplied  by  the  slant  height. 

2.  Show  that  the  formula  for  computing  the  lateral  area  of  a  right  cir- 
cular cone  is  obtained  from  the  formula  for  the  lateral  area  of  a  frustum 
of  a  right  circular  cone  by  making  one  of  the  bases  of  the  frustum  reduce 
to  a  point. 

3.  Write  the  formula  for  the  total  area  of  a  frustum  of  a  right  circu- 
lar cone. 

4.  If  a  trapezoid,  one  of  whose  non-parallel  sides  is  perpendicular  to 
the  bases,  is  revolved  about  this  side  as  axis,  the  lateral  area  of  a  frus- 
tum of  a  cone  traced  equals  2  tt  times  the  product  of  the  other  non-parallel 
side  and  the  line-segment  joining  the  middle  points  of  the  non-parallel 
sides. 

Suggestion.  —  See  Exercise  1  above. 

5.  The  slant  height  of  a  frustum  of  a  right  circular  cone  is  14  in. 
and  the  diameters  of  the  bases  are  16  in.  and  10  in.,  respectively.  Find 
the  lateral  area.     The  total  area. 

6.  The  altitude  of  a  frustum  of  a  right  circular  cone  is  8  ft.,  and  the 
radii  of  the  bases  4  ft.  and  9  ft.,  respectively.  Find  the  slant  height.  The 
lateral  area.     The  total  area. 

7.  A  funnel  is  made  of  tin,  in  which  the  diameter 
AB  is  8  in.,  the  diameter  CD  is  1  in.,  the  diameter  EF  is 

I  in.,  the  length  ^C  is  5  in.,  and  the  length  CE  is  4  in. 
Find  the  amount  of  tin  required  to  make  it,  not  allowing 
for  the  seams. 

8.  Find  the  amount  of  sheet  metal  required  to  make 
a  lot  of  1000  pails,  each  10  in.  deep,  8  in.  in  diameter  at  the  bottom,  and 

II  in.  in   diameter  at  the  top,  not  allowing  for  seams   nor  waste  in 
cutting. 

9.  Given  the  diameters  and  the  altitude  of  a  frustum  of  a  right 
circular  cone,  explain  how  to  draw  the  pattern  by  which  a  piece  of  sheet 
metal  must  be  cut  out  in  order  to  be  rolled  up  and  made  into  the  re- 
quired frustum. 

Suggestions.  —  Find  the  slant  height.  Let  x  =  the  radius  of  the 
inner  arc  of  the  pattern.     Then  x  +  the  slant  height  =  the  radius  of  the 


382  SOLID   GEOMETRY 

outer  arc.     These  radii  are  proportional  to  the  circumferences  of  the 
bases  of  the  frustrum.     Solve  the  proportion  for  x. 

10.  A  tinner  wishes  to  make  a  coifee  pot  which  shall  be  7  in.  deep, 
4  in.  in  diameter  at  the  top,  and  6  in.  in  diameter  at  the  bottom.  Al- 
lowing a  quarter  of  an  inch  for  a  seam,  cut  from  paper  a  pattern  for  the 
conical  surface. 

Note.  —  For  explanation  of  the  terms  used,  and  of  the  process  in- 
volved in  the  manufacture  of  articles  as  indicated  in  the  following  prob- 
lems, see  Problem  9,  page  352. 

11.  A  stew  pan  is  to  be  made  in  the  form  of  a  frustum  of  a  right  cir- 
cular cone,  6^  in.  wide  at  the  bottom, 

8^  in.  wide  at  the  top,  3  in.  deep,  and 
to  have  a  flange  \  in.  wide  at  the  top. 
Find  the  diameter  of  the  blank  from 
which  it  must  be  made. 

Suggestion.  —  The  area  of  the  pan  consists  of  three  parts :  the 
bottom  ;  the  side,  which  is  the  lateral  area  of  a  frustum  of  a  cone  ;  and 
the  flange,  which  is  the  diif erence  between  the  areas  of  two  circles.  The 
sum  of  these  must  equal  the  area  of  the  blank.  Indicate  the  factors,  and 
avoid  as  much  multiplication  as  possible. 

12.  A  tin  dish  pan  is  to  be  made  13  in.  wide  at  the  bottom,  17  in. 
wide  at  the  top,  6  in.  deep,  and  have  a  ^-in.  flange  at 'the  top.  Find  the 
diameter  of  the  blank  from  which  it  must  be  made. 

13.  A  tin  pie  pan  is  6f  in.  in  diameter  at  the  bottom,  8  in.  in  diameter 
at  the  top,  1  in.  deep,  and  has  a  ^-in.  flange  at  the  top.  Find  the  diameter 
of  the  blank  from  which  it  is  made. 

14.  A  safety  valve  of  an  engine  has  a  valve  V  in  the  form  of  a  frus- 
tum of  a  cone,  which  closes  the  opening  D  in 
the  boiler  AB.  The  diameter  of  the  opening 
Z>  is  2  in.  The  conical  surface  of  the  valve  is 
inclined  at  45°.  The  "  effective  area,"  or  area 
of  the  opening  through  which  the  steam  escapes 
when  the  valve  V  is  lifted,  is  evidently  the  lateral  area  of  a  frustum  of  a 
cone  whose  slant  height  is  ss. 

If  the  valve  V  is  lifted  through  a  height  of  \  in.,  find  the  effective 
area. 


PYRAMIDS  AND  CONES  383 

413.    Theorem.  —  The  volume  of  a  circular  cone  is  equal  to  one 
third  of  the  product  of  its  altitude  and  the  area  of  its  base. 


Hypothesis.  F"=  volume,  H=  altitude,  and  B  =  the  area 
of  the  base  of  a  circular  cone. 

Conclusion.      F'=  J  HB. 

Suggestions.  Inscribe  a  pyramid  whose  base  is  a  regular 
polygon. 

What  formula  expresses  the  volume  of  the  pyramid  ? 

The  formula  V=\  HB  may  be  proved  from  this  by  §  275, 
(1),  if  \  HB  is  first  proved  to  be  the  limit  of  what  ? 

Write  the  proof  in  full. 

414.  Corollary.  —  If  Vis  the  volume  of  a  circular  cone^  H  the 
altitude^  and  R  the  radius  of  the  hase^  then 

V^iirB^H, 
The  proof  is  left  to  the  student. 

EXERCISES 

1.  The  vohime  of  a  circular  cone  is  one  third  of  the  volume  of  a 
circular  cylinder  having  the  same  base  and  altitude. 

2.  Prove  the  theorem  in  §  413  by  circumscribing  about  the  cone  a 
pyramid  with  a  regular  base. 

3.  Find  the  volume  of  a  circular  cone  of  which  the  altitude  is  16  in. 
and  the  diameter  of  the  base  12  in. 

4.  Find  the  volume  of  a  right  circular  cone  the  radius  of  whose  base 
is  6  in.  and  whose  slant  height  is  20  in. 


384 


SOLID   GEOMETRY 


5.  A  fruit  raiser  has  a  round  pile  of  apples  that  is  8  ft.  across  at  the 
bottom,  and  tapers  to  a  point  5  ft.  high  at  the  middle.  Allowing  3  bu.  to 
4  cu.  ft.,  how  many  bushels  are  there  in  the  pile? 

6.  A  farmer  has  a  pile  of  ear  corn  approximately  in  the  form  of  a 
cone  whose  height  is  10  ft.  and  width  at  the  bottom  20  ft.  Allowing 
2  bu.to  5  cu.  ft.,  how  many  bushels  does  it  contain? 

7.  Crushed  stone  falling  from  a  stone  crusher  forms  a  round  pile  30 
ft.  around  at  the  bottom  and  5  ft.  high  at  the  center.  How  many  cubic 
yards  of  crushed  stone  are  there  in  the  pile  ? 

8.  A  farmer  wishes  to  know  how  many  tons  of  hay  there  are  in  a 
stack  in  the  form  of  a  cylinder  18  ft.  in  diameter  and  8  ft.  high,  sur- 
mounted by  a  cone  10  ft.  high.  Allowing  512  cu.  ft.  to  the  ton,  find  the 
amount  of  hay  in  the  stack. 

9.  A  farmer  has  a  rick  of  hay  36  ft.  long  and  16  ft.  wide.  The 
lower  part  is  made  up  of  a  parallel- 
opiped  with  half  cylinders  at  its 
ends,  and  is  8  ft.  high.  The  part 
surmounting  this  tapers  to  a  line 
20  ft.  long  and  18  ft.  above  the 
ground,  and  is  made  up  of  a  trian- 
gular prism  with  half  cones  at  its 
ends.     Find  the  number  of  tons  in  it,  allowing  512  cu.  ft.  to  the  ton. 

10.  A  hill,  approximately  in  the  form  of  a  cone,  is  to  be  removed  and 
the  ground  leveled  down  even  with  its  base,  and  laid  off  into  building 
lots.  The  diameter  of  the  base  is  280  ft.,  and  the  highest  point  is  12  ft. 
above  the  level.  How  many  loads  (cubic  yards)  of  earth  must  be  re- 
moved ? 

11.  In  the  center  of  a  building  lot  is  a  piece  of  elevated  ground 
approximately  in  the  form  of  a  cone.  The  elevated  part  is  to  be  cut 
down  and  used  to  raise  the  level  of  the  surrounding  ground.  The  lot  is 
100  ft.  wide  and  120  ft.  long.  The  height  of  the  cone  is  5  ft.  and  the 
diameter  of  its  base  60  ft.  Find  how  much  the  elevated  part  must  be 
cut,  and  how  much  the  rest  of  the  lot  must  be  filled,  to  bring  it  to  a  level. 

12.  Given  the  lateral  area  5  of  a  right  circular  cone,  and  the  radius 
R  of  the  base.     Find  the  volume  V. 

13.  Given  the  total  area  T  of  a  right  circular  cone,  and  the  lateral  area 
S.    Find  the  volume  V, 


PYRAMIDS  AND  CONES  385 

415.  Theorem.  —  The  volume  of  a  fruBtum  of  a  circular  cone 
is  equal  to  one  third  of  the  product  of  its  altitude  and  the  sum  of 
the  areas  of  the  bases  and  the  mean  proportional  between  them. 


Hypothesis.  V=  volume,  11=  altitude,  and  B  and  b  respec- 
tively =  the  areas  of  the  bases  of  a  frustum  of  a  circular  cone. 

Conclusion.      F  =  ^  H(iB  +  6  +  V56) . 

Suggestions.  Inscribe  a  frustum  of  a  pyramid  whose  base 
is  a  regular  polygon. 

What  formula  expresses  the  volume  of  the  frustum  of  a 
pyramid  ? 

The  formula  V=\ H(B  +  6  -f- ^Bb)  may  be  proved  from 
this  by  §  275,  (1),  if  ^H{B  +  5  +  V^)  is  first  proved  to  be 
the  limit  of  what  ? 

Write  the  proof  in  full. 

416.  Corollary.  —  If  Vis  the  volume  of  a  frustum  of  a  circu- 
lar cone^  H  its  altitude^  and  R  and  r  respectively  the  radii  of 
the  bases^  then 

V=\7rH(iB?  +  r^+  Rry 

The  proof  is  left  to  the  student. 

EXERCISES 

1.  The  volume  of  a  frustum  of  a  circular  cone  is  equal  to  the  sum  of 
the  volumes  of  the  three  circular  cones  having  a  common  altitude  equal 
to  the  altitude  of  the  frustum,  and  for  bases  the  two  bases  of  the  frustum 
and  a  mean  proportional  between  them,  respectively. 

2.  Find  the  volume  of  a  frustum  of  a  circular  cone  of  which  the  alti- 
tude is  4  in.  and  the  diameters  of  the  bases  6  in.  and  5  in.,  respectively. 


386 


SOLID   GEOMETRY 


3.  Find  the  volume  of  a  frustum  of  a  right  circular  cone  of  which 
the  slant  height  is  32  in.  and  the  radii  of  the  bases  16  in.  and  28  in. 
respectively. 

4.  Does  a  liquid  measure  with  a  diameter  of  8  in. 
at  the  top,  a  diameter  of  13 1  in.  at  the  bottom,  and 
12|  in.  deep,  contain  5  gal.?     (1  gal.  =  231  cu.  in.) 

5.  A  Sanford's  Ink  bottle  is  in  the  form  of  a  frus- 
tum of  a  cone  whose  height  is  If  in.,  and  the  diame- 
ters at  the  top  and  bottom  1^  in.  and  2  in.,  respectively.     How  many  of 
these  bottles  can  be  filled  with  a  gallon  of  ink  ? 

6.  The  base  of  a  marble  column  is  a  frustum  of  a  cone.  The  height 
is  1  ft.  6  in.,  and  the  diameters  of  the  bases  6  ft.  and  5  ft.  6  in.,  respec- 
tively.    Allowing  170  lb.  to  the  cubic  foot,  find  its  weight. 

7.  A  Dutch  windmill,  in  the  shape  of  a 
frustum  of  a  cone,  is  30  ft.  high.  The  outer 
diameters  at  the  bottom  and  top  are  15  ft.  and 
12  ft.,  respectively;  and  the  inner  diameters  at 
the  bottom  and  top  are  12  ft.  and  9  ft.,  respec- 
tively. How  many  cubic  yards  of  stone  were 
required  to  build  it  ? 

8.  The  chimney  of  a  factory  is  to  be  100  ft. 
high ;  the  outer  diameters  at  the  bottom  and 
top  are  to  be  18  ft.  and  10  ft.,  respectively ;  and 
the  flue  is  to  be  6  ft.  in  diameter  throughout. 
How  many  thousands  of  bricks,  2  in.  by  4  in. 

by  8  in.,  must  the  contractor  order  for  it,  allowing  10%  for  mortar? 

9.  A  churn,  in  the  form  of  a  frustum  of  a  cone,  is  28  in.  high,  10  in. 
in  diameter  at  the  bottom,  and  8  in.  in  diameter  at  the  top.  How  many 
gallons  of  cream  will  it  hold? 

10.  A  cake  pan  is  10  in.  in  diameter  at  the  top 
and  8  in.  in  diameter  at  the  bottom,  and  its  depth 
is  3  in.  Rising  from  the  center  of  the  bottom  is  a 
stem  in  the  form,  of  a  frustum  of  a  cone  1^  in.  in 
diameter  at  the  bottom  and  1  in.  in  diameter  at  the 

top,  which  makes  the  hole  in  the  cake.     How  many  cubic  inches  does  the 
pan  hold  ? 


PYRAMIDS  AND   CONES 


387 


417.  Theorem.  —  The  lateral  or  the  total  areas  of  two  similar 
cones  of  revolution  are  to  each  other  as  the  squares  of  the  slant 
heights^  the  altitudes^  or  the  radii  of  the  bases  ;  and  their  vol- 
umes are  to  each  other  as  the  cubes  of  the  slant  heights^  the  alti- 
tudes^ or  the  radii  of  the  bases. 


Hypothesis.  S  and  s  are  the  lateral  areas,  T  and  t  the 
total  areas,  V  and  v  the  volumes,  L  and  I  the  slant  heights, 
IT  and  h  the  altitudes,  and  R  and  r  the  radii  of  the  bases,  of 
two  similar  cones  of  revolution. 

Conclusion.      -  =  ^  =  _  =  _=_;    _  =  _  =  _  =  _. 

Proof.  The  proof  is  left  to  the  student.  Proceed  as  in 
§  379  and  §  382.     Write  the  proof  in  full. 


EXERCISES 

1.  The  lateral  area  of  a  cone  of  revolution  is  144  sq.  in.  What  is  the 
lateral  area  of  a  similar  cone  whose  altitude  is  f  that  of  the  given  cone  ? 

2.  The  altitude  of  a  circular  cone  is  8  ft.  How  far  from  the  vertex 
must  a  plane  be  passed  parallel  to  the  base  to  cut  off  -^^  of  the  cone  from 
the  vertex  ? 

3.  To  divide  a  circular  oone  into  halves  by  a  plane  parallel  to  the  base, 
how  far  from  the  vertex  must  the  plane  be  passed  ? 

4.  From  a  cone  of  revolution  of  which  the  slant  height  is  24  ft.  and 
»f  which  the  diameter  of  the  base  is  12  ft.,  a  cone  of  which  the  slant 
height  is  8  ft.  is  cut  off  by  a  plane  parallel  to  the  base.  Find  the  lateral 
area  and  the  volume  of  the  cone  cut  off;  of  the  frustum.  -       •^- 


388 


SOLID   GEOMETRY 


MISCELLANEOUS  EXERCISES 

1.  A  lamp  shade,  made  of  a  wire  frame  covered  with  silk,  is  in  the 
form  of  a  frustum  of  a  right  circular  cone  whose  slant 

height  is  9  in.  and  the  radii  of  whose  bases  are  6  in.  and 
4^  in.,  respectively.  If  one  half  extra  is  allowed  for 
fullness  in  shirring  the  silk,  how  much  silk  is  required 
for  the  lamp?  If  the  silk  is  27  in.  wide,  find  to  an 
eighth  of  a  yard  how  much  of  a  yard  is  required. 

2.  The  lateral  area  of  a  right  circular  cone  is  twice 
the  area  of  the  base.  Compare  the  slant  height  and  the 
radius.  At  what  angle  does  an  element  of  the  cone 
meet  the  plane  of  the  base  ? 

3.  If  r  is  the  radius  of  the  base  of  a  right  circular  cone  and  h  the 
slant  height,  prove  that  the  ratio  of  the  area  of  the  base  to  the  lateral 


area  is  -  • 
h 

4.  If  a  right  circular  cone  and  a  right  circular  cylinder  have  equal 
bases  and  the  slant  height  of  the  cone  is  equal  to  the  altitude  of  the 
cylinder,  prove  that  the  entire  area  of  the  cylinder  is  equal  to  twice  the 
entire  area  of  the  cone. 

5.  An  ice  cream  dipper  is  in  the  form  of  a  right 
circular  cone  whose  diameter  is  2f  in.  and  altitude 
2f  in.     It  is  filled   level   full   at  each  serving.     How 

many  servings  can  be  obtained  from  a  gallon  (231  cu.  in.)  of  ice  cream  ? 
What  is  received  for  a  gallon  at  5  ^  a  serving  ? 

6.  Triangle  ABC  is  an  isosceles  triangle 
with  equal  sides  AB  and  AC.  Line  DE  is 
drawn  through  A  parallel  to  J5C,  and  A  ABC  is 
revolved  about  DE  as  an  axis.  Find  the  vol- 
ume of  the  figure  traced  by  A  ABC. 

7.  A   conical  glass  beaker,  graduated    for 
measuring  liquids,   holds   a  quart.      At  what 
point  of  the  scale  must  the  mark  be  placed  to  measure  a  pint? 
measure  a  gill  ? 

8.  A  farmer  has  two  similar  piles  of  corn,  each  in  the  form  of  a 
right  circular  cone.  The  diameters  of  the  two  piles  are  4  ft.  and  18  ft., 
respectively.     He  measures  the  small  pile,  and  finds  that  it  contains 


PYRAMIDS  AND   CONES 


389 


10  bu.     How  can  he  estimate,  without  measuring,  the  number  of  bushels 
in  the  large  pile?     How  many  bushels  are  there  in  the  large  pile  ? 

9.  Two  boards,  A  and  B,  are  placed  in  parallel  vertical  positions  on 
a  table.  A  has  a  square  opening  cut  in  it.  A  lamp  is  placed  at  S. 
Light  from  this  lamp  passes  through  the 
opening  in  A,  and  illuminates  part  of  B. 
Show,  by  aid  of  this  diagram,  that  the 
intensity  of  the  light  falling  upon  a  given 
surface  is  inversely  proportional  to  the 
square  of  the  distance  of  the  surface  from 
the  source  of  light. 

10.  The  altitude  of  each  of  two  pyramids 
whose  bases  are  in  the  same  plane  is  16  ft. 
The  base  of  one  is  a  square  whose  side  is 
12  ft.,  and  the  base  of  the  other  is  a  regular  hexagon  whose  side  is  8  ft. 
The  area  of  the  section  of  the  first  made  by  a  plane  parallel  to  the  base 
is  81  sq.  ft.  What  is  the  area  of  the  other  pyramid  made  by  the  same 
plane  ? 

11.  A  fruit  vender  has  a  load  of  apples  in  a  wagon  bed  that  is  3  ft. 
wide,  1  ft.  deep,  and  10  ft.  long.  The  apples  fill  the  bed  and  are  piled 
up  in  approximately  a  pyramid  to  a  point  1  ft.  higher  than  the  top  of 
the  bed.     How  many  bushels  are  there  ?     (Count  3  bu.  to  4  cu.  ft.) 

12.  A  triangular  piece  of  ground  ABC  is  to  be  leveled.  The  side 
AB  is  level,  and  the  corner  C  is  12  ft.  higher  than  AB.  Find  how  much 
the  ground  must  be  cut  down  at  C  and  how  much  it  must  be  fiUed  along 
^^  to  make  it  level. 

13.  Find  the  volume  of  a  regular  quadrangular  pyramid  whose 
altitude  is  20  ft.  and  slant  height  28  ft. 

14.  Find  the  volume  of  a  triangular  pyramid  each  edge  of  which  is 

4  ft. 

15.  The  slant  height  of  a  frustum  of  a  regular  quadrangular  pyramid 
is  18  in.,  and  the  sides  of  the  bases  are  20  in.  and  12  in.,  respectively. 
Find  its  volume. 

16.  The  lateral  edge  of  a  frustum  of  a  regular  hexagonal  pyramid  is 
6  ft.,  and  the  sides  of  the  bases  are  8  ft.  and  4  ft.,  respectively.  Find  its 
volume. 


CHAPTER   XV 

REGULAR   POLYEDRONS.     SIMILAR   POLYEDRONS. 
PRISMATOIDS 

418.  Regular  polyedrons.  —  A  regular  polyedron  is  a  poly- 
edron  all  of  whose  faces  are  congruent  regular  polygons  and 
all  of  whose  polyedral  angles  are  equal. 

Models  of  five  regular  polyedrons  are  as  follows.  That 
these  are  the  only  kinds  possible  is  proved  in  §  419. 


Tetraedron        Hexaedron       Octaedron        Uodecaedron       Icosaedron 

419.  Theorem.  —  There  cannot  he  more  than  jive  kinds  of 
regular  polyedrons. 

Proof.  1.  Each  angle  of  an  equilateral  triangle  is  60°;  of 
a  square,  90°;  of  a  regular  pentagon,  108°;  etc.  §  101 

2.  .*.  not  more  than  three  kinds  of  regular  polyedrons  are 
possible  with  equilateral  triangles  as  faces  (tetraedron^  octa- 
edron^ icosaedron') ;  for  3  x  60°,  4  x  60°,  or  5  x  60°  is  less  than 
360°,  while  6  x  60°  or  any  greater  multiple  of  60°  is  not  less 
than  360°.  §  343 

3.  Similarly,  not  more  than  one  kind  of  regular  polyedron 
is  possible  with  squares  as  faces  (hexaedron  or  cube);  for 
3  X  90°  is  less  than  360°,  while  4  x  90°  or  any  greater  multiple 
of  90°  is  not  less  than  360°. 

390 


REGULAR  POLYEDRONS,   ETC. 


391 


4.  Similarl}^  not  more  than  one  kind  of  regular  polyedron 
is  possible  with  regular  pentagons  as  faces  (dodecaedron) . 

5.  No  regular  polyedron  is  possible  with  regular  hexagons 
or  regular  polygons  of  more  sides  as  faces,  because  the  sum 
of  the  face  angles  at  each  vertex  would  not  be  less  than  360°. 

6.  .-.  only  five  regular  polyedrons  are  possible. 

Note.  —  The  regular  polyedrons  were  studied  by  the  ancient  Greeks. 
The  members  of  the  Pythagorean  school  knew  that  there  were  five  of  these 
solids.  It  is  said  that  Hippasus  (about  470  e.g.),  a  Pythagorean  who  dis- 
covered the  regular  dodecaedron,  was  drowned  for  announcing  his  discovery, 
because  the  Pythagoreans  were  a  secret  society,  and  the  members  were 
pledged  to  refer  the  honor  of  any  new  discovery  to  Pythagofas,  the  founder. 

420.  Construction  of  models  of  regular  polyedrons.  —  Models 
of  the  five  regular  polyedrons  may  be  constructed  of  card- 
board by  first  drawing  patterns  as  shown  below,  then  cutting 
out  the  patterns,  folding  along  the  dotted  lines,  and  pasting. 


,  r 

1  j 

1/ 

1 

1 

V — 4 

r 

EXERCISES 

1.  How  many  faces  has  a  regular  tetraedron?     Hexaedron?    Octa- 
@dron  ?     Dodecaedron  ?     Icosaedron  ? 

2.  Construct  cardboard  models  of  one  or  more  of  the  five   regular 
polyedrons  as  suggested  in  §  420. 

3.  Prove  that  a  regular  hexaedron  has  a  center,  i.e.  a  point  within 
equidistant  from  all  vertices.     Is  this  true  of  all  regular  polyedrons  ? 


392  SOLID   GEOMETRY 

421.  Similiar  polyedrons.  —  Two  polyedrons  are  similar  when 
they  have  the  same  number  of  faces,  similar  each  to  each 
and  similarly  placed,  and  the  homologous  polyedral  angles 
are  equal. 


Similar  Polyedrons 

422.   Corollary.  —  In  two  similar  polyedrons : 

(1)  The  ratio  of  any  two  homologous  edges  is  equal  to  the 
ratio  of  any  other  two  homologous  edges. 

(2)  The  areas  of  any  two  homologous  faces  are  to  each  other 
as  the  squares  of  any  two  homologous  edges. 

(3)  The  areas  of  the  entire  surfaces  are  to  each  other  as  the 
squares  of  any  two  homologous  edges. 

The  proof  is  left  to  the  student. 

EXERCISES 

1.  In  two  similar  polyedrons,  any  two  homologous  diagonals  have  the 
same  ratio  as  any  two  homologous  edges. 

2.  Prove  that  any  two  regular  tetraedrons  are  similar.  Is  this  true 
of  any  two  regular  hexaedrons?  Octaedrons?  Dodecaedrons  ?  Icosae- 
drons  ? 

3.  Two  homologous  edges  of  two  similar  polyedrons  are  4  in.  and  6  in., 
respectively.  The  total  area  of  the  smaller  is  128  sq.  in.  Find  the  total 
area  of  the  larger. 

4.  Verify  §  422  (3),  in  the  case  of  two  similar  rectangular  parallelo- 
pipeds  whose  concurrent  edges  are  3  in.,  4  in.,  6  in.,  and  6  in.,  8  in., 
12  in.,  respectively. 


REGULAR  POLYEDRONS,   ETC. 


393 


423.  Theorem.  —  Two  tetraedrons  are  similar  when  the  faces 
including  a  triedral  angle  of  one  are  similar^  respectively^  to  the 
faces  including  a  triedral  angle  of  the  other^  and  are  similarly 
'placed. 


Hypothesis.     Tetraedrons  ABOD  and  EFGRh^iwe 

AABC'^AEFG,  AACD'^AJSaff,  AABB'^AEFH. 
Conclusion.     AB  CD  ~  EFGH. 

Proof.     1.   AABO'^AEFG,   AACD^AEaH,  A  ABB 
~  A  EFH.  Hyp. 


3. 


BO 
Fa 
BO 

Fa 


4.    Similarly, 

5. 
6. 


AO 

Ea 

OB 

an' 

OB 


and 


OB 

aR' 


AO 

Ea 


§127 


Ax.  I 


^BB 

aH     FE' 

.\  A  BOB ^  A  Fan.  §129 

Now  Z  BAO  =  Z  FEa,  Z.  OAB  =  Z  aEH,  Z  BAB 

=  Z  FEE,  §  127 

7.  .-.  angle  A-BOB  =  angle  E-FaH.  §  346 

8.  Similarly,  the  other  corresponding  triedral  angles  are 
equal. 

9.  :.ABOB^EFaH.  §421 

EXERCISE 
If  a  giyen  tetraedron  is  cut  by  a  plane  parallel  to  a  face,  the  tetra- 
edron  cut  off  is  similar  to  the  given  one. 


394  SOLID   GEOMETRY 

424.    Theorem.  —  Two  tetraedrons  are  similar  when  a  diedral 
angle  of  one  is  equal  to  a  diedral  angle  of  the  other,  and  the 
including  faces  are  similar  each  to  each  and  similarly  placed. 
A 


Hypothesis.  In  tetraedrons  ABQD  and  EFGiH,  diedral 
angle  C-AB-D  =  diedral  angle  G-UF-H,  A  ABC^AJEFa, 
AABB^AFFH. 

Conclusion.     AB  OB  ~  FFaH. 

Proof.     1.    AnglQ  C-AB-D  = 'dugiQ  a-FF-H,  QtQ.     Hyp. 

2.  .-.  ABCB  may  be  superposed  upon  FFG-H  so  that 
angle  0-AB-B  coincides  with  its  equal,  angle  Gr-FF-H, 
A  falling  at  F.  Test  of  equality 

3.  ZBA0=AFFa3indZBAB  =  ZFFH.  §127 

4.  .-.  J. (7  will  fall  along  FG,  and  AD  along  Fff.  §  13 

5.  .-.  Z  CAD^^Z  GFR.  §  13 

a    AC     AB       .  AD     AB  «  .07 

6.  -^^-^=^^7:^  and  -——  =  ——•  §  127 
Fa     FF          FH     FF  ^ 


Ax.  I 


7    .  A^^A^ 

'     ' Fa~ FH 

8.  ..AACDr^AFaS,  §130 

9.  .'.ABCD^FFaH,  §423 

EXERCISE 
Is    this    a   true   theorem:    Tetraedron    ^5CZ>  ~  tetraedron    EFGH 
when  AABC-^AEFG,  angle   C-^5-Z>  =  angle  G-EF-H,  and  angle 
B-AC-D  =  angle  F-EG-Hl 


REGULAR  POLYEDRONS,   ETC. 


395 


425.  Theorem.  —  Two  similar  polyedrons  may  he  decomposed 
into  the  same  number  of  tetraedrons^  similar  each  to  each  and 
similarly  placed. 


Hypothesis.     AM  audi  UN"  are  two  similar  polyedrons. 

Conclusion.  AM  and  ^iV  may  be  decomposed  into  the  same 
number  of  tetraedrons,  similar  each  to  each  and  similarly 
placed. 

Proof.     1.    Polyedrons  J-iHf  and  JS'iV  are  similar.  Hyp. 

2.  Let  A  and  ^  be  any  two  homologous  vertices.  Divide 
all  faces,  except  those  having  A  and  U  as  vertices,  into  tri- 
angles. Pass  planes  through  A  and  the  vertices  of  the 
triangles  in  the  faces  of  AM^  and  through  U  and  the  vertices 
of  the  triangles  in  the  faces  of  UI^. 

3.  Let  ABQB  and  EFG-H  be  two  homologous  tetra- 
edrons formed. 

4.  Now  face  AB  ~  face  EF. 

BO^AO 

Fa    Fa' 

6.  r.AABC^AFFa, 

7.  Similarly,  AACB'^A  EaS. 

8.  Also  angle  B-AC-D  =  angle  F-Ea-H. 

9.  ,'.ABCD^EFaH. 
10.  It  may  now  be  proved  that  tetraedrou  ABDM 

edron  EFHN,  etc. 


Z  BOA  =  Z  FaE,  and  ^  =  ^ 


§421 
§127 
§130 

§421 
§424 
tetra- 


396  SOLID   GEOMETRY 

426.  Theorem.  —  The  volumes  of  two  tetraedrons  ha-^xng  a 
pair  of  equal  triedral  angles  are  to  each  other  as  the  products 
of  the  edges  including  the  equal  triedral  angles. 


E 


A 


Hypothesis.  Tetraedrons  ABCD  and  UFG^R  hsive  angle 
B-A  CD  =  angle  F-UGrff,  and  their  volumes  are  V  and  W, 
respectively. 

^     ,    .  r     BAxBOxBI) 

Conclusion.       -:fj>  =  -:prp, ^pr?^ 777^. 

W      FE  X  Fa  X  FH 
Proof.     1.    Angle  ^-^(7i>  =  angle  J^-^(5^ir.  Hyp. 

2.  .*.  ABOD  may  be  superposed  upon  FFGrH  so  that 
angles  B-AOD  and  F-FGrH  coincide,  ABOD  becoming 
OFPQ,  Test  of  equality 

3.  If  Oilif  and  ^iV^are  J.  plane  FaH,  OMW  EK       §  321 

4.  .-.  Oil[f  and  ^i\r  determine  a  plane,  which  contains  FE 
and  intersects  plane  FGrRin  FN,  §  299,  §  291,  §  301 

.      V_iOMxAFPQ^OM    AFFQ  .004 

'  w  ^ENxAFas   EN    A  Fas'  ^ 

6.  In  AFMO  and  FNE,  Z  F  is  common,  Z  FMO  = 
Z  FNE,  Z  FOM=  FEN.  §  26 

7.  .'.AFMO^AFNE.  §128 

8.  ,,OM^FO  ^^  BA^ 

EN     FE         FE  ^ 

9.  Also  ^^^  =  -^^x^g  or  ^Ox_BB^  228 

A  Fas     FGxFH        FaxFH  ^ 

10         V ^BA     BO  xBD^BAxBOxBD      .      ^^j 
'   "W    FE     FaxFH    FEx  FaxFH       ^' 


REGULAR  POLYEDRONS,  ETC.  397 

427.    Theorem.  —  The  volumes  of  two  similar  tetraedrons  are 
to  each  other  as  the  cubes  of  their  homologous  edges. 


Hypothesis.  Tetraedrons  ABOD  and  EFG-H  are  similar, 
with  AB  and  EF  homologous  edges,  and  with  volumes  V 
and  W^  respectively. 

^     ,    .         V     Aff 

Concmsion.   ^=v=-=t;« 
W    EF^ 

Itr^^i.     1.    Tetraedron^J^CZ^-'tetraedron  ^^(5^j?.   Hyp. 
2.    .-.  angle  A-BCD  =  angle  E-FGH.  §  421 

3     .  V^ABy^AC^AB^AB     AQ     AT)  «  .^fi 

'   "W     EFxEGxEH     EF     Ea     EH'  ^ 

4.    But  — =  — =  — .  §422a) 

EF     Ea     EH  ^         ^^ 

c         V     AB^AB     AB       Aff  .      VTT 

W     EF     EF     EF       EF^ 

428.    Corollary. —  The  volumes  of  any  two  similar  polyedrons 
are  to  each  other  as  the  cubes  of  their  homologous  edges. 
The  proof  is  left  to  the  student. 


EXERCISES 

1.  The  edge  of  a  cube  is  16  in.     Find  the  edge  of  a  cube  which  shall 
have  a  volume  twice  as  great  as  that  of  the  given  cube. 

2.  If  the  edge  of  a  given  cube  is  e,  find  the  edge  of  a  cube  with  n 
times  the  volume. 


398  SOLID   GEOMETRY 

3.  The  edge  of  a  pyramid  is  x.  Find  the  edge  of  a  similar  pyramid 
with  twice  the  volume. 

4.  The  edge  of  a  given  polyedron  is  E.  Find  the  edge  of  a  similar 
polyedron  with  m  times  the  volume. 

5.  A  rectangular  tank  is  4  ft.  by  6  ft.  by  8  ft.  What  are  the  dimen- 
sions of  a  similar  tank  that  will  hold  4  times  as  much? 

6.  The  volumes  of  two  similar  polyedrons  are  36  cu.  in.  and  972  cu. 
in.,  respectively.  The  edge  of  the  smaller  is  3  in.  Find  the  homologous 
edge  of  the  larger. 

7.  A  farmer  had  two  rectangular  ricks  of  hay  of  the  same  shape. 
The  large  rick  was  1\  times  as  long  as  the  small  one.  He  hauled  away 
and  weighed  the  small  one,  and  found  that  it  contained  2\  toL<s.  How 
many  tons  did  the  large  rick  contain  ? 

8.  A  farmer  had  two  similar  wedge-shaped  piles  of  corn.  One  was 
10  ft.  high  and  the  other  only  4  ft.  high.  The  small  pile  was  known  to 
contain  70  bu.     How  many  bushels  in  the  larger  pile? 

9.  The  volume  of  a  given  octaedron  is  72  cu.  in.  Construct  a  similar 
octaedron  whose  volume  shall  be  9  cu.  in. 

10.  The  altitude  of  a  pyramid  is  9  ft.  Find  where  to  pass  two  planes 
through  it,  parallel  to  the  base,  that  will  cut  it  into  three  equal  parts. 

11.  From  a  given  pyramid  cut  off  a  frustum  whose  volume  shall  be  | 
of  the  volume  of  the  given  pyramid. 

12.  The  volumes  of  two  similar  prisms  are  64  cu.  ft.  and  1000  cu.  ft., 
respectively.  The  total  area  of  the  smaller  is  32  sq.  ft.  Find  the  total 
area  of  the  larger. 

13.  If  S  and  s  are  the  total  areas,  and  V  and  v  the  volumes  of  two  sim- 
ilar polyedrons,  write  the  equation  expressing  the  relation  between  them. 

429.  Prismatoids.  —  A  prismatoid  is  a  polyedron  bounded 
by  two  polygons  in  parallel  planes,  called  the  bases,  and  by 
lateral  faces  which  are  triangles,  parallelograms,  or  trape- 
zoids having  one  side  common  with  one  of  the  bases  and  the 
opposite  vertex  or  side  common  with  the  other  base. 

The  perpendicular  distance  between  the  planes  of  the  bases 
is  called  the  altitude.  The  section  made  by  a  plane  parallel  to 
the  bases  and  bisecting  the  altitude  is  called  the  mid-section. 


REGULAR  POLYEDRONS,  ETC. 


399 


430.  Theorem.  —  The  volume  of  any  prismatoid  is  equal  to 
one  sixth  of  the  product  of  its  altitude  hy  the  sum  of  the  areas 
of  its  bases  and  four  times  its  mid-section. 

A 


Hypothesis.  F'is  the  volume  of  a  prismatoid,  JJ  the  alti- 
tude, B  and  h  areas  of  bases,  and  il[f  area  of  mid-section. 

Conclusion.      V=^\  HiB  +  h +  ^  M). 

Proof.  1.  F=  volume,  5"=  altitude,  B  and  5  =  areas 
of  bases,  M=  area  of  mid-section.  Hyp. 

2.  Let  0  be  any  point  in  mid-section.  Pass  a  plane 
through  0  and  each  edge  of  the  prismatoid,  dividing  it  into 
pyramids  with  common  vertex  0, 

3.  The  volumes  of  the  pyramids  whose  bases  are  B  and  h 
are  \  HB  and  \  Hh^  respectively.  §  395 

4.  All  pyramids  having  lateral  faces  of  the  prismatoid  as 
bases  may  be  divided  into  triangular  pyramids,  as  0-ABC, 

5.  It  may  be  proved  that  A  ABC  =4:  x  A  AEF.  §  228 
.-.  0-ABC  =4  X  0-AEF.  Ax.  IV,  §  394 
But  0-AEF  =  A-  OEF  =  i  J2^  x  A  OEF,  §  394 
.-.  0-^^(7  =  1  ^x  (4  X  A  OEF^,  Ax.  XII 
In  adding  all  such  pyramids  as  0-ABC^  the  sum  of 

all  such  triangles  as  A  OEF  would  equal  M.  Ax.  X 

10.  .-.  the  sum  of  all  such  pyramids  =  \H  y^^lM. 

Ax.  II,  Factoring,  Ax.  XII 

11 .  .-.  V^\  E^B -t-  6  -h 4  Jf ).  Ax.  II,  Factoring 


6. 
7. 
8. 
9. 


400 


SOLID   GEOMETRY 


Note.  —  The  formula  F=  |  H{B  +  6  +  4  ilf )  for  computing  the  volume 
of  any  prismatoid  is  very  important.  From  it  can  be  derived  the  formulae 
for  computing  the  volumes  of  all  of  the  solids  of  elementary  geometry.  One 
illustration  of  this  is  shown  in  the  following  corollary. 

431.  Corollary.  —  The  volume  of  any  truncated  triangular 
prism  is  equal  to  the  product  of  the  area  of  a  right  section  and 
one  third  of  the  sum  of  the  three  lateral  edges. 

For,  if  g,  /,  and  g  are  the  lateral 
edges  and  h  the  altitude  and  h  the  base 
of  a  right  section,  by  considering  the 
edge  e  as  one  base  and  the  opposite 
lateral  face  as  the  other  base  of  a 
prismatoid. 

But  M=\hil[e+n+l[e  +  g-]), 

=  ihbx\(,e+f  +  g^ 
=  area  of  right  section  x  ^  (^  -f-/-}-  ^). 
Explain  the  steps  of  the  proof. 


EXERCISES 

1.  Show  that  the  rule  in  §  370  for  finding  the  volume  of  any  prism 
follows  from  the  prismatoid  formula  by  making  B  =  h  =  M. 

2.  Show  that  the  rule  in  §  395  for  finding  the  volume  of  any  pyra- 
mid follows  from  the  prismatoid  formula. 

3.  Show  that  the  rule  in  §  397  for  finding  the  volume  of  a  frustum 
of  a  pyramid  follows  from  the  prismatoid  formula. 

Suggestions.  —  1.  If  S  and  s  are  corresponding  sides  of  B,  and  5,  then 
the  corresponding  side  of  il/  is  |  {S  +  s). 


2.   Hence,  show  that 


By  adding  these  equations  and  solving,  show  that 


REGULAR  POLYEDRONS,   ETC. 


401 


4.  A  prismatoid  of  which  one  base  is 
a  rectangle,  called  the  lase^  and  the  other 
base  a  line  parallel  to  one  side  of  this 
rectangle,  called  the  edge,  is  a  wedge. 

Develop  the  formula  for  the  volume  of 
a  wedge. 

5.  The  base  of  a  wedge  is  5  in.  by  8  in.,  the  altitude  6  in.,  and  the 
edge  4  in.     Find  the  volume. 

6.  Show  by  use  of  the  prismatoid  formula  that  the  volume  of  any 
truncated  quadrangular  prism  whose  opposite  lateral  faces  are  parallel 
equals  the  area  of  a  right  section,  nmltiplied  by  one  fourth  of  the  sum 
of  the  four  lateral  edges. 

7.  A  truncated  right  quadrangular  prism  has  for  base  a  square  whose 
side  is  3  in.,  and  its  lateral  edges  are  4  in.,  5  in.,  4  in.,  and  3  in.,  respect- 
ively.     Find  its  volume. 

8.  A  marble  monument  is  in  the  form  of  a  truncated  right  quadran- 
gular prism  whose  base  is  a  rectangle  1  ft.  6  in.  by  3  ft.,  and  lateral  edges 
3  ft.,  3  ft.,  3^  ft.,  and  3^  ft.,  respectively.  Marble  weighs  170  lb.  to  the 
cubic  foot.     Find  the  weight  of  the  monument. 

9.  A  bin  of  wheat  is  8  ft.  wide  and  10  ft.  long.  The  depths  of  the 
wheat  at  the  four  corners  are  4  ft.,  6  ft.,  6  ft.,  and  8  ft.,  respectively. 
Find  the  number  of  bushels  in  the  bin.     (Count  4  bu.  to  5  cu.  ft.) 

10.  In  finding  the  contents  of  large  excavations, 
surveyors  lay  out  the  surface  of  the  ground  to  be 
excavated,  such  as  AEOMRP,  into  equal  rectan- 
gles. Pegs  are  driven  at  the  corners  of  all  the 
rectangles,  s.uch  as  G,  H,  etc.,  and  the  depth  of  the 
cut  to  be  made  at  each  of  these  corners  is  found 
with  a  surveyor's  level.  For  practical  purposes, 
when  the  rectangles  are  small,  the  surface  of  any  one  rectangle  may  be 
considered  plane.  The  whole  excavation  is  thus  divided  into  a  number 
of  partial  volumes,  each  in  the  form  of  a  truncated  quadrangular  prism. 
By  computing  the  volume  of  each  of  these,  and  adding,  the  volume  of 
the  whole  excavation  is  found.  Evidently,  the  depth  of  the  cut  at  any 
corner  may  be  used  1,  2,  3,  or  4  times  in  the  computation,  depending 
upon  the  number  of  rectangles  adjoining  it. 

Show  that  the  volume  of  the  whole  excavation  may  be  found  by  the 
following  rule : 


A 

B      C 

D       E 

F 
K 

Lj_    M 

! 

N\      0 

P 

(^     R 

402  SOLID   GEOMETRY 

"  Take  each  corner  height  as  many  times  as  there  are  partial  areas 
adjoining  it,  add  them  all  together,  and  multiply  by  one  fourth  of  the 
area  of  a  single  rectangle." 

11.  In  the  figure  of  Exercise  10,  each  rectangle  is  a  square  whose  side 
is  50  ft.     The  depths  of  the  cuts  at  the  various  corners  are  as  follows 
A,  1  ft. ;  B,  6  ft. ;  C,  3  ft. ;  I),  4  ft. ;  E,  4  ft. ;  F,  3  ft. ;  G,  5  ft. ;  H,  1  ft. 
/,3  ft.;  J,  3  ft.;  X,  2  ft.;  X,  4  ft. ;  ikf,  3  ft.;  iV,  5  ft.;  0,4  ft.;  P,  3  ft. 
Q,  5  ft. ;  R,  6  ft.    Find  the  number  of  cubic  feet  in  the  excavation.    The 
number  of  cubic  yards. 

12.  In  the  excavation  of  Wash-        K  *  c   I 

ington  Street,  between  1st  Street     -J  =^  i^-J- £ — £ — S-Ii  ^  L 

and  2nd  Street,  stakes  are  driven     -,  ^      WASHINGTON  ST      ^ 
every  100  ft.  on  each  side  of  the        ^^  ]^G    H      I       J      K    L^      ^ 
street.     The  width  of  "Washington 

Street  is  50  ft.  The  depths  of  the  cuts  are  found  to  be  as  follows: 
^,8ft.;  A6ift.;  C,4ft.;  A  5  f  t. ;  E,2ft.;  F,lft.;  (?,7ft.;  iy,7ft.; 
/,  5i  ft. ;  J,  4  ft. ;  K,  3^  ft. ;  L,  2  ft.  Find  the  number  of  cubic  yards 
in  the  excavation. 

MISCELLANEOUS  BXEROISBS 

1.  Two  polyedrons  which  can  be  divided  into  the  same  number  of 
tetraedrons,  similar  each  to  each  and  similarly  placed,  are  similar. 

2.  The  middle  points  of  the  edges  of  a  regular  tetraedron  are  the 
vertices  of  a  regular  octaedron. 

3.  If  from  any  point  within  a  regular  tetraedron  perpendiculars  are 
drawn  to  the  faces,  their  sum  is  equal  to  an  altitude  of  the  tetraedron. 

4.  Construct  an  angle  equal  to  the  plane  angle  of  a  diedral  angle  of 
a  regular  octaedron. 

5.  The  volunie  of  a  regular  octaedron  is  equal  to  the  product  of  the 
cube  of  its  edge  and  J- V2. 

6.  If  each  of  two  polyedrons  is  similar  to  a  third  polyedron,  they  are 
similar  to  each  other. 

7.  The  homologous- edges  of  three  similar  tetraedrons  are  3,  4,  and  5, 
respectively.  What  is  the  homologous  edge  of  a  similar  tetraedron  whose 
volume  is  equal  to  the  sum  of  the  volumes  of  the  three  tetraedrons? 

8.  If  two  polyedrons  ABODE'-'  and  UKLM--  are  so  placed 
that  line-segments  from  a  point   0  to  A^  B,  C,  et,}.,  are  divided  by 


REGULAR  POLYEDRONS,  ETC. 


403 


points  /,  J",  Kf  etc.,  such  that 

OA^OB^qC 
01      OJ     OK 


^-etc 
OL-^^'' 


the  two  polyedrons  are  said 
to  be  radially  placed.  Prove 
that  polyedrons  ABODE  ••• 
B.ndIJKLM  •"  are  similar. 

9.  A  plane  passed  through 
the  point  of  intersection  of  the 
diagonals  of  any  parallelopiped  divides  it  into  two  equal  solids. 

10.  The  volume  of  a  truncated  parallelopiped  is  equal  to  the  area  of  a 
right  section,  multiplied  by  the  distance  between  the  points  of  intersec- 
tion of  the  diagonals  of  the  bases  (centers). 

11.  Find  the  area  and  the  volume  of  a  regular  tetraedron  each  edge  of 
which  is  4  in. 

12.  The  total  area  of  a  given  tetraedron  is  256  sq.  ft.  Find  the  total 
area  of  the  tetraedron  cut  oif  by  a  plane  parallel  to  a  face  of  the  given 
tetraedron  and  midway  between  that  face  and  the  opposite  vertex. 


CHAPTER  XVI 

THE  SPHERE 

432.  A  sphere.  —  A  sphere  is  a  solid  bounded  by  a  curved 
surface  all  points  of  which  are  equally  distant  from  a  point 
within  called  the  center.  The  bounding  surface  is  called  the 
spherical  surface ;  the  line-segment  from  the  center  to  any 
point  of  the  spherical  surface,  a  radius ;  and  a  line-segment 
through  the  center  and  terminating  in  the  spherical  surface, 
a  diameter. 

As  in  a  circle,  a  diameter  of  a  sphere  is  equal  to  two  radii. 


A  sphere  may  be  traced  by  revolving  a  semicircle  through 
one  revolution  about  its  diameter  as  an  axis,  for  all  points  of 
the  surface  thus  formed  are  equally  distant  from  the  center 
of  the  semicircle. 

Thus,  by  revolving  the  semicircle  A  CB  through  one  revolution  about 
its  diameter  AB  as  an  axis,  the  sphere  with  center  O  and  diameter  CD 
is  traced. 

433.  Fundamental  properties  of  spheres.  —  The  following  im- 
portant properties  of  spheres  may  easily  be  reasoned  out  from 
the  definitions  above ; 

404 


''^  THE  SPHERE  405 

(1)  All  radii  of  the  same  sphere  or  of  equal  spheres  are 
equal. 

(2)  All  diameters  of  the  same  sphere  or  of  equal  spheres 
are  equal. 

(3)  Two  spheres  of  equal  radii.,  or  of  equal  diameters^  are 
equal, 

(4)  The  distance  of  a  point  from  the  center  of  a  sphere  is 
equal  to.,  greater  than.,  or  less  than  a  radius  according  as  it  is 
on.,  without,  or  within  the  spherical  surface;  and  conversely. 

434.  Theorem.  —  A  section  of  the  surface  of  a  sphere  made 
by  a  plane  is  a  circle. 


Hypothesis.  ABO  is  a  section  of  the  surface  of  a  sphere 
with  center  0  made  by  a  plane. 

Conclusion.     ABQ  is  a  circle. 

Proof.  1.  ABO  is  a  section  of  the  surface  of  a  sphere 
with  center  0  made  by  a  plane. 

2.  Draw  CD  ±  plane  of  ABO,  meeting  it  at  D.  Let  A 
and  ^  be  any  two  points  in  the  section.  Driaw  OA.,  OB,  DA, 
and  BB. 

3.  Then  OA  =  OB.  §  433,  (1) 

4.  .-.  DA  =  DB.  §  311 

5.  Therefore,  since  A  and  B  are  any  two  points  in  section 
ABO,  all  points  of  the  section  are  equidistant  from  D,  and 
hence  ABO  is  a  circle.  Def.  O 


406  SOLID   GEOMETRY 

435.  Circles  of  a  sphere.  —  The  section  of  the  surface  of  a 
sphere  made  by  a  plane  passing  through  the  center  is  called  a 
great  circle  of  the  sphere. 

The  section  of  a  surface  of  a  sphere  made  by  a  plane  which 
does  not  pass  through  the  center  is  called  a  small  circle  of 
the  sphere. 

The  following  are  important  properties  of  the  circles  of  a 
sphere,  and  should  be  carefully  reasoned  out  by  the  student : 

(1)  All  great  circles  of  a  sphere  are  equal. 

(2)  Tlie  plane  of  any  great  circle  bisects  the  sphere  and  its 
surface. 

(3)  Two  great  circles  of  a  sphere  bisect  each  other. 

(4)  Two  points  in  the  surface  of  a  sphere  which  are  not  ex- 
tremities of  a  diameter  determine  a  great  circle  of  the  sphere. 

(5)  Any  three  points  in  a  surface  of  a  sphere  determine  a 
circle  of  the  sphere. 

EXERCISES 

1.  What  are  the  meridians  on  the  earth's  surface  ?  How  are  they 
determined  ? 

2.  What  are  the  lines  called  parallels  on  the  earth's  surface  ?  Why 
are  they  so  named  ? 

3.  The  sky  overhead  appears  to  be  a  great  spherical  dome.  Why  does 
a  telescope,  when  revolved  in  one  plane 

about  an  axis,  appear  to  describe  a  circle 
across  the  sky  ?  The  drawing  shows  the 
great  telescope  in  the  Yerkes  Astronomi- 
cal Observatory  at  Lake  Geneva,  Wis. 

4.  The  line  joining  the  center  of  a 
sphere  and  the  center  of  a  small  circle  of 
the  sphere  is  perpendicular  to  the  plane 
of  the  circle. 

5.  The  radius  of  a  sphere  is  20 
in.  Find  the  radius  and  area  of  the 
circle  made  by  a  plane  4  in.  from  the 
center. 


THE    SPHERE 


407 


6.  At  what  distance  from  the  center  of  a  sphere  16  in.  in  diameter 
should  a  plane  be  passed  to  produce  a  circle  whose  diameter  is  12  in.  ? 

7.  Circles  of  a  sphere  made  by  planes  equidistant  from  the  center 
are  equal. 

8.  State  and  prove  the  converse  of  Exercise  7. 

9.  Of  two  circles  on  the  same  sphere,  the  one  made  by  a  plane  more 
remote  from  the  center  is  the  smaller. 

10.  State  and  prove  the  converse  of  Exercise  9. 

11.  What  is  the  largest  number  of  points  in  which  two  circles  of  a 
sphere  can  intersect?    Why? 

12.  The  intersection  of  the  surfaces  of  two   spheres   is    a   circle 
whose  center  is  on  the  line-segment  joining  the  centers  of  the  spheres 
and  whose  plane  is  perpendicular  to  the  line- 
segment. 

Suggestion.  —  Let  P  and  Q  be  the  centers 
of  the  two  spheres,  and  let  a  plane  through  PQ 
cut  the  surfaces  of  the  spheres  in  great  circles 
which  intersect  at  A  and  B.  Prove  that  by 
revolving  the  figure  about  PQ  as  axis,  the  cir- 
cles describe  the  surfaces  of  the  spheres  and  the  point  A  describes  a  circle 
which  is  their  intersection. 

13.  The  radii  of  two  intersecting  spheres  are  8  in.  and  10  in.,  respec- 
tively. The  distance  between  their  centers  is  12  in.  Find  the  radius 
and  the  area  of  their  circle  of  intersection. 

14.  The  stars  appear  to  lie  upon  the  surface  of  a  sphere,  called  the 
astronomical  sphere^  with  the  earth  at  its  center.  Because  of  the  daily 
rotation  of  the  earth  on  its  axis,  all  stars,  except  the 

Pole  Star,  appear  to  move.  Since  the  distance  from 
any  star  to  the  Pole  Star  appears  to  be  always  the 
same,  show  that  the  apparent  daily  path  of  a  star  is  a 
circle. 

Suggestion.  —  If  P  is  the  position  of  the  Pole 
Star,  E  of  the  earth,  and  S  of  any  other  star,  ES 
and  Z  SEP  are  constant.     Draw  SC  ±  PE, 


408  SOLID   GEOMETRY 

436.    Axis  and  poles  of  a  circle.  —  The  axis 

of  a  circle  of  a  sphere  is  the  diameter  of  the 
sphere  which  is  perpendicular  to  the  plane  of 
the  circle.  V  / 

The  poles  of  a  circle  of  a  sphere  are  the  ex-       \^^^/ 
tremities  of  the  axis  of  the  circle. 


m 


Thus,  if  AB  is  the  diameter  of  a  sphere  which  is  perpendicular  to  the 
plane  of  a  circle  of  the  sphere  at  C,  ^^  is  the  axis  of  the  circle  and 
^  and  ^  are  its  poles. 

It  follows  from  the  proof  in  §  434  that: 

The  axis  of  a  circle  of  a  sphere  passes  through  the  center  of 
the  circle. 

EXERCISES 

1.  Considering  the  earth  as  a  sphere,  the  North  Pole  and  South  Pole 
are  really  poles  of  what  circles  ? 

2.  A  great  circle  of  a  sphere  which  passes  through  one  pole  of  a  circle 
must  pass  through  the  other  pole  also. 

3.  Circles  of  a  sphere  made  by  parallel  planes  have  the  same  axis 
and  the  same  poles. 

4.  A  given  point  on  the  surface  of  a  sphere  is  the  pole  of  one  and 
only  one  great  circle. 

5.  The  line-segment  which  is  perpendicular  to  the  plane  of  a  circle  of 
a  sphere  at  the  center  of  the  circle  and  terminates  in  the  surface  of  the 
sphere  is  the  axis  of  the  circle. 

Suggestion.  —  This  may  be  proved  if  it  is  first  shown  that  the  given 
line-segment  coincides  with  the  axis.    Hence  begin  by  drawing  the  axis. 

437.  Distance  on  the  surface  of  a  sphere.  —  The  distance  be- 
tween any  two  points  on  the  surface  of  a  sphere  is  the  length 
of  the  smaller  arc  of  the  great  circle  passing  through  the 
points.  If  the  two  arcs  are  equal,  either  may  be  taken  to 
represent  the  distance. 


THE   SPHERE 


409 


438.    Theorem.  — Either  pole  of  a  circle  of  a  sphere  is  equi- 
distant from  all  points  on  the  circle. 

P 


Hypothesis.  A  and  B  are  any  two  points  on  circle  ABC^ 
P  and  Q  are  the  poles  of  circle  ABC^  and  AP,  BP^  J^,  and 
BQ  are  arcs  of  great  circles. 

Conclusion.     AP  =  BP  and  AQ  =  BQ. 

Proof.  1.  A  and  B  are  any  two  points  on  circle  ABC  of 
which  P  and  Q  are  the  poles.  Hyp. 

2.  Let  axis  PQ  meet  the  plane  of  ABO  Rt  D,  the  center 
of  the  circle.     Draw  AD,  BD,  and  chords  AP  and  BP. 

3.  P§±  plane  of  ABQ.  §  436 

4.  AD  =  i52).  §  151,  (2) 

5.  .-.  AP  =  BP.  §  310 

6.  .'.AP  =  BP.  ^  §151,  (7) 

7.  Similarly  it  may  be  proved  that  AQ  =  BQ. 

439.  Polar  distance.  —  The  polar  distance  of  a  circle  of  a 
sphere  is  the  distance  from  any  point  of  the  circle  to  the 
nearer  pole.  Thus,  in  the  figure  of  §  438,  AP  is  the  polar 
distance  of  circle  ABC. 

440.  Corollary. —  The  polar  distance  of  a  great  circle  is  a 
quadrant. 

The  proof  is  left  to  the  student. 


410  SOLID   GEOMETRY 

441.  Theorem.  — If  a  'point  on  the  surface  of  a  sphere  is  at 
a  quadrant's  distance  from  each  of  two  other  points  on  the 
surface  which  are  not  extremities  of  a  diameter,  it  is  a  pole  of 
the  great  circle  determined  hy  those  points. 


Hypothesis.  P,  A,  and  B  are  points  on  the  surface  of  the 
sphere  with  center  0  ;  AP  and  BP  are  quadrants;  ABQ  is 
the  great  circle  determined  by  A  and  B, 

Conclusion.     P  is  the  pole  of  ABO. 

Suggestions.  It  will  follow  that  P  is  the  pole  of  ABCii 
it  is  proved  that  PO  1.  plane  of  ABC,  Why?  The  latter 
will  follow  if  what  is  first  proved  ?  Hence  begin  by  drawing 
AG,  BO,  and  PO,  and  proving  Z  AOP  and  Z.  BOP  right 
angles. 

EXERCISES 

1.  Prove  that  if  the  distance  between  two  points  on  a  sphere  is  a 
quadrant,  each  point  is  the  pole  of  one  great  circle  through  the  other. 

2.  Prove  that  if  ^  is  a  point  on  the  surface  of 
a  material  sphere,  globe,  or  spherical  blackboard, 
a  circle  BCD  with  A  as  its  pole  may  be  drawn 
on  the  surface  by  holding  one  end  of  a  piece  of 
cord  at  A,  inserting  a  pencil  or  crayon  in  a  loop 
at  the  other  end  of  the  cord,  and  marking  along 
the  surface  while  the  cord  is  kept  stretched. 
Compasses  may  be  used  instead  of  the  cord. 

3.  Prove  that  the  diameter  of  a  given  material 
sphere  may  be  obtained  with  straightedge  and 
compasses  as  follows : 

With  any  point  P  of  the  surface  as  pole,  draw  any  circle  ABC, 


THE  SPHERE 


411 


With  compasses,  measure  the  chords  joining  any  three  points  A,  5, 
and   C  of  this  cir-  p 

cle,  and  construct  a  ..  M 

plane  triangle  DEF 
with  its  sides  equal 
to  these  chords. 

Circumscribe  a 
circle  about  A  DEF, 
and  let  0  be  its 
center. 

Construct  right  A  MNQ  with  hypotenuse  MN  =  chord  PC,  and  leg 
QN"  =  radius  OF,  and  complete  right  A  MNR. 

Then  MR  is  the  diameter  required. 

442.  Lines  and  planes  tangent  to  a  sphere.  —  A  straight  line 
or  a  plane  is  tangent  to  a  sphere  if  it  has  in  common  with  the 
surface  of  the  sphere  one  and  only  one  point,  called  the  point 
of  contact. 

443.  Theorem.  —  A  plane  perpendicular  to  a  radius  of  a 
sphere  at  its  extremity  is  tangent  to  the  sphere* 


Hypothesis.  0  is  the  center  of  a  sphere,  OA  is  a  radius, 
and  plane  MN  is  perpendicular  to  OA  at  A. 

Conclusion.     MNi^  tangent  to  the  sphere  with  center  0. 

Suggestions.  It  may  be  proved  that  MNis  tangent  to  the 
sphere  if  it  is  shown  that  any  point  other  than  A  of  plane 
MN,  such  as  B,  is  outside  of  the  sphere.  Why  ?  This  will 
follow  if  it  is  first  proved  that  OB  >  OA.     Why  ? 

Write  the  proof  in  full. 


412  SOLID   GEOMETRY 

444.  Theorem. — A  plane  tangent  to  a  sphere  is  perpendiculat 
to  the  radius  drawn  to  the  point  of  contact. 

Suggestions.  Let  plane  MN  be  tangent  at  -4  to  a  sphere 
with  center  0.  I^et  B  be  any  point  of  iifiV  except  A.  Show- 
that  OB  >  OA,  etc. 

Write  the  proof  in  full. 

EXERCISES 

1.  A  perpendicular  to  a  tangent  plane  at  the  point  of  tangency  passes 
through  the  center  of  the  sphere. 

2.  A  line  perpendicular  to  a  tangent  plane  from  the  center  of  a 
sphere  passes  through  the  point  of  contact. 

3.  Two  planes  tangent  to  a  sphere  at  the  extremities  of  the  same 
diameter  are  parallel. 

4.  Show  that  the  diameter  of  any  solid  spherical  ball  may  be  found 
by  laying  the  ball  upon  a  table,  resting  a  board  upon  it  in  a  horizontal 
position,  and  measuring  the  perpendicular  dis- 
tance from  the  edge  of  the  board  to  the  table. 

Suggestion.  —  Draw  a  diameter  perpendicu- 
lar to  the  table  top. 

5.  A  straight  line  perpendicular  to  the  radius 
of  a  sphere  at  its  extremity  is  tangent  to  the 
sphere. 

6.  A  straight  line  tangent  to  a  sphere  is  per- 
pendicular to  the  radius  drawn  to  the  point  of 
contact. 

7.  All  straight  lines  tangent  to  a  sphere  at  the  same  point  lie  in  the 
plane  which  is  tangent  to  the  sphere  at  that  point. 

445.  Circumscribed  and  inscribed  spheres  of  a  polyedron.  —  A 
sphere  is  circumscribed  about  a  polyedron  when  all  of  the  ver- 
tices of  the  polyedron  lie  in  the  surface  of  the  sphere.  The 
polyedron  is  said  to  be  inscribed  in  the  sphere. 

A  sphere  is  inscribed  in  a  polyedron  when  all  of  the  faces  of 
the  polyedron  are  tangrent  to  the  sphere.  The  polyedron  is 
said  to  be  circumscribed  about  the  sphere. 


THE   SPHERE 


413 


446.     Theorem.  —  A  sphere  can  he  circumscribed  about  any 
tetraedron. 


Hypothesis.    ABCD  is  any  tetraedron. 

Conclusion.     A  sphere  can  be  circumscribed  about  ABOD. 

Proof.     1.    ABOD  is  any  tetraedron.  Hyp. 

2.  Let  M  and  N  be  the  centers  of  the  circumscribed 
circles  of  A  ACD  and  A  BQD^  respectively. 

3.  Then,  iHf  and  iV^lie  in  perpendicular  bisectors  ME  and 
NE  of  CD.  §  105 

4.  .-.  plane  MEN  JL  line  CD.  §  303 
.-.  plane  MEN  1.  planes  ACD  and  BCD.  §  331 
Let  MF be  A.  plane  ACD  and  NG  be  A.  plane  BCD. 
Then  MF  and  iV^^  both  lie  in  plane  MEN.  §  383 
.'.  ilO^and  NG-  intersect  at  some  point  0.  §  55 
0  is  equidistant  from  A.,  (7,  and  i),  and  also  equidis- 
tant from  B,  C,  and  D.                                                        §  310 

10.  .-.  0  is  equidistant  from  J.,  ^,  (7,  and  D.  Ax.  I 

11.  ,*.  the  surface  of  a  sphere  with  center  0  and  radius  OA 
will  pass  through  J.,  B^  (7,  and  i>.  Def .  sphere 

12.  .*.  this  sphere  is  circumscribed  about  ABCD.       §  445 

EXERCISES 

1.  A  sphere  can  be  circumscribed  about  any  rectangular  parallelo- 
piped. 

2.  A  sphere  can  be  inscribed  in  any  cube. 


5. 

6. 

7. 
8. 
9. 


414  SOLID   GEOMETRY 

447.  Principles  of  limits.  — If  one  half  of  a  regular  polygon 
having  an  even  number  of  sides  is  inscribed  in  a  semicircle^  and 
the  whole  figure  is  revolved  about  the  diameter  of  the  semicircle 
as  axis^  then: 

(1)  the  area  of  the  surface  of  the  sphere  traced  is  the  limit 
approached  by  the  area  of  the  surface  traced  by  the  perimeter 
of  the  half  polygon  as  the  number  of  sides  of  the  polygon  is 
indefinitely  increased; 

(2)  the  volume  of  the  sphere  traced  is  the  limit  approached 
by  the  volume  of  the  solid  traced  by  the  half  polygon  as  the 
number  of  sides  of  the  polygon  is  indefinitely  increased. 

448.  Theorem.  —  The  area  of  the  surface  of  a  sphere  is  equal 
to  four  times  the  area  of  a  great  circle  of  the  sphere. 


Hypothesis.  S  is  the  area  of  the  surface  and  M  the  radius 
of  the  sphere  with  center  (9,  traced  by  revolving  semicircle 
ACF about  diameter  AF as  axis. 

Conclusion.     S=4:'jrB^. 

Proof.  1.  AS'=area  of  surface,  jB=  radius,  0  is  center  of 
sphere  traced  by  semicircle  AOF  revolved  about  diameter 
AF  as  axis.  Hyp. 

2.  Inscribe  in  the  semicircle  one  half  of  a  regular  poly- 
gon of  an  even  number  of  sides,  ABO-",     Draw  OM 1.  any 


Def.  sim.  A 


THE  SPHERE  415 

side,  as  BC.     From  B,  M,  and   Q  draw  JiBG;  etc.,  to  AF. 
And  draw  BK±  OH, 

3.  In  revolving  trapezoid  BCHGi  about  AF  as  axis,  BO 
will  trace  the  lateral  surface  of  a  frustum  of  a  right  circular 
cone  whose  area  equals  ir  x  BO  x  {BQ- -\-  OH),  §  412 

4.  But  il[f  is  the  middle  point  of  ^(7.  §156 

5.  Ba,  MN,  and  (Tff  are  parallel.  §  41 

6.  .•.MN=^(iBa+OH^.  §100 

7.  .-.  lateral  area  of  frustum  traced  hj B0=2ir  x  BO  x 
MK  Ax.  XII 

8.  It  may  be  proved  that  A  0KB  ~  A  MNO.  (Proof 
left  to  student.) 

Q     .    OM^MJSr 
"BO     BK 

10.  But  BK=  aH,  §  85 

11.  ,,OM^m.  Ax.  XII 

BO    an 

12.  .',  OMxaB^BOxMN.  Ax.  IV 

13.  .-.  area  of  surface  traced  by  BO  =2  ir  x  OM x  GrH. 

Ax.  XII 

14.  Also  it  can  be  proved  that 

area  of  surface  traced  by  AB  =  lTr  x  OM x  AG-^ 
area  of  surface  traced  by  OD  =  2  tt  x  OM  x  HL^  etc. 

15.  .*.  if  B  represents  the  area  of  the  surface  traced  by 
ABO"',  «  =  27rx  0MxAa+27rx  OMx  aB:-\-27rx  OMx 
HL  +  etc.  Ax.  II 

16.  .'.s=^2'jr  X  OMxiAa^  GH^-HL+  Qic.^,      Fact. 

17.  .\8  =  2irx  0Mx2B,  Ax.  XII 

18.  If  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  8  =  aS'  and  0M=  R.  §  447,  §  274 

19.  .'.27rxOMx2B=2'7rxRx2Rov4'7rB^. 

§  275,  (2) 

20.  .\S=^7rB^.  §  275,  (1) 


416  SOLID   GEOMETRY 

449.  Corollary  l.  —  The  areas  of  the  surfaces  of  two  spheres  are 
to  each  other  as  the  squares  of  their  radii  or  of  their  diameters. 

The  proof  is  left  to  the  student. 

450.  Zones.  —  A  zone  is  that  portion  of  the  surface  of  a  sphere 
which  lies  between  two  parallel  planes. 


/    •-«^f«-.-^..  Zone 

The  sections  of  the  surface  made  by  the  planes  are  the 
bases  of  the  zone.  The  perpendicular  distance  between  the 
planes  is  the  altitude  of  the  zone. 

If  one  of  the  planes  is  tangent  to  the  sphere,  the  zone  is 
called  a  zone  of  one  base. 

When  a  semicircle  is  revolved  about  the  diameter  as  an 
axis,  the  surface  traced  by  any  arc  of  the  semicircle  is  a  zone. 

Thus,  in  the  figure  of  §  448,  BD  traces  a  zone,  and  GL  is  the  altitude. 

451.  Corollary  2. —  The  area  of  a  zone  of  a  sphere  is  equal 
to  the  product  of  its  altitude  ayid  the  circumference  of  a  great 
circle. 

For,  the  method  of  proof  in  §  448  applies  equally  to  the 
zone  traced  by  an  arc  of  the  revolving  semicircle.  The  sides 
of  the  inscribed  polygon  which  are  inscribed  in  the  given 
arc  trace  surfaces  whose  areas  are  expressed  as  in  step  14  of 
§  448.  By  adding  these  areas  and  finding  the  limit  of  their 
sum,  the  equation  aS'  =  2  irRH  is  obtained,  where  S  is  the 
area  and  H  the  altitude  of  the  zone,  and  B  the  radius  of  the 
sphere. 


THE   SPHERE 


417 


EXERCISES 

1.  Considering  the  entire  surface  of  a  sphere  as  a  zone,  deduce  the 
formula  for  computing  the  area  of  the  surface  of  a  sphere  from  §  451. 

2.  Show  that  the  area  of  the  surface  of  a  sphere  is  found  by  the 
formula  S  =  ttD^  where  D  is  the  diameter. 

3.  What  is  the  area  of  the  surface  of  a  sphere  whose  diameter  is  10  in.  ? 

4.  The  area  of  the  surface  of  a  sphere  is  equal  to  the  product  of  the 
diameter  and  the  circumference  of  a  great  circle. 

5.  The  diameter  of  a  given  sphere  is  12  in.  If  parallel  planes  divide 
the  diameter  into  four  equal  parts,  find  the  area  of  each  of  the  four  zones 
thus  cut  off. 

6.  Zones  on  the  same  sphere  or  on  equal  spheres  are  to  each  other 
as  their  altitudes. 

7.  The  diameter  of  a  sphere  is  8  in.,  and  a  plane  is  passed. through 
the  sphere  1  in.  from  the  center.  Compare  the  two  parts  into  which  the 
surface  of  the  sphere  is  divided. 

8.  Find  the  diameter  of  a  sphere  the  area  of  whose  surface  is  9C  sq.  ft. 

9.  The  figure  represents  a  sphere  inscribed  in  a  right  circular  cylinder, 
the  bases  of  the  cylinder  being  tangent  to  the  sphere  and  the  lateral  sur- 
face of  the  cylinder  touching  the  surface  of  the 
sphere  in  a  circle  of  contact.  Corresponding 
belts  are  formed  on  the  two  surfaces,  included 
between  two  planes  each  parallel  to  the  bases  of 
the  cylinder.  Prove  that  the  areas  of  the  two 
belts  are  equal.  Show  that  the  entire  area  of 
the  surface  of  the  sphere  is  equal  to  the  lateral 
area  of  the  cylinder. 

10.  The  area  of  the  surface  of  a  sphere  is 
equal  to  two  thirds  of  the  total  area  of  the  surface 
of  the  circumscribed  right  circular  cylinder. 

Note.  —  The  discovery  of  the  relation  in  Exercise  10  between  the  areas  of 
the  surfaces  of  a  sphere  and  its  circumscribed  cylinder,  and  of  a  like  relation 
between  the  volumes  of  these  solids,  was  first  made  by  Archimedes.  He  was 
so  pleased  with  these  discoveries  that  he  asked  that  the  figure  of  a  sphere 
and  its  circumscribed  cylinder  be  inscribed  on  his  tomb.  This  request  was 
carried  out  by  his  friend  Marcellus. 


418  SOLID    GEOMETRY 

11..  Prove  that  a  zone  of  one  base  has  the  same  area  as  a  cir«le  whose 
radius  is  the  chord  of  the  generating  arc  of  the  zone. 

Suggestion.  —  This  necessitates  showing  that  the  square  of  the  chord 
is  equal  to  2  times  the  product  of  the  radius  and  the  altitude  of  the 
zone.     Why? 

12.  Taking  the  earth  as  a  sphere  and  the  radius  as  4000  miles,  find 
the  area  of  the  entire  surface  of  the  earth.  The  area  of  the  United  States 
is  3,624,122  sq.  mi.  What  part  of  the  earth's  surface  is  occupied  by  the 
United  States? 

13.  The  North  Frigid  Zone  is  that  part 
of  the  earth's  surface  lying  north  of  the 
Arctic  Circle  and  extending  to  the  North 
Pole.  Its  altitude  is  approximately  330  mi. 
Find  its  area. 

What  part  of  the  earth's  surface  lies  in 
the  two  frigid  zones? 

14.  Find  the  diameter  of  the  Arctic 
Circle. 

15.  The  North  Temperate  Zone  is  that  part  of  the  earth's  surface  lying 
between  the  Tropic  of  Cancer  and  the  Arctic  Circle.  Its  altitude  is  ap- 
proximately 2165  mi.     Find  its  area. 

What  part  of  the  earth's  surface  lies  in  the  two  temperate  zones? 

16.  Prove  that  one  half  of  the  earth's  surface  lies  within  30°  of  the 
equator. 

Suggestion.  —  What  is  the  relation  between  the  sides  of  a  right 
triangle  in  which  one  of  the  acute  angles  is  30°  ? 

17.  How  many  miles  above  the  surface  of  the  earth  is  that  point  from 
which  one  fourth  of  the  earth's  surface  may  be  seen? 

Suggestion.  —  Apply  the  same  principles  as  in  Exercise  16. 

18.  An  observer  stands  d  feet  from  the  center  of  a  sphere  whose 
diameter  is  d  feet.  What  part  of  the  surface  of  the  sphere  can  be 
seen? 

19.  Find  the  total  area  of  a  hemispherical  bowl  1  in.  thick  whose 
external  diameter  is  12  in. 

Suggestion.  —  We  are  to  find  the  surfaces  of  two  hemispheres  and  9 
ring  (the  rim). 


THE   SPHERE  419 

20.  The  surface  of  a  tiled  dome,  in  the  form  of  a  hemispherical  sur- 
face whose  diameter  is  24  ft.,  is  made  of  colored  tiles  each  1  in.  square. 
How  many  tiles  are  required  to  make  it? 

21.  In  a  zone  of  one  base,  the  diameter  of  the  base  is  16  in.  and  the 
altitude  4  in.     Find  the  radius  of  the  sphere. 

Suggestion.  —  If  x  is  the  radius  of  the  sphere,  4(2  a:  —  4)  =  64. 

22.  Vessels  whose  surfaces  are  in  the  form  of  zones  of  one  base  some- 
times are  made    from   sheet 

metal,  by  cutting  out  a  cir- 
cular blank  from  a  flat  sheet 
of  the  metal,  and  pressing  it 
into  the  required  form  in  a 
die,  as  described  in  the  exer-  ""^^^ilPtes^ 

cises  on  the  cylinder  and  the 

cone.     Find  the  diameter  of  the  blank  required  to  make  a  hemispherical 
brass  bowl  whose  diameter  is  8  in. 

23.  The  lid  to  a  silver-plated  dish  is  in 
the  form  of  a  zone  of  one  base,  with  a 
flange.  The  depth  of  the  lid  is  1^  in., 
the  width  of  the  flange  1  in.,  and  the  outer 
diameter  of  the  flange  9  in.  Find  the  diameter  of  the  blank  from  which 
it  is  made. 

24.  Hollow  metal  balls,  used  as  ornaments,  casters,  anti-friction  balls, 
etc.,  are  made  from  circular  blanks  cut  from  sheet  metal,  by  use  of  a  die. 
Find  the  diameter  of  the  blank  required  to  make  a  hollow  metal  ball 
^  in.  in  diameter. 

25.  A  method  of  testing  the  hardness  of  metals  consists  in  partly 
forcing  a  hardened  steel  ball  into  the  sample  being  tested.  The  liard- 
ness  numeral  H  is  computed   by  the  formula 

H  =  —,  where  K  is  pressure  on  ball  in  kilo- 

y 

gi  ams,  and  y  is  area  of  surface  of  depression  in 
sample  in  square  millimeters. 

Show  that  y  =  2  irr{r  —  Vr^  -  BPi,  where  r  — 
radius  of  ball  and  U  —  radius  of  depression  in 
sample. 

Substitute  this  value  of  y,  and  deduce  the  formula  for  finding  H  in 
terms  of  K^  r,  and  R. 


420 


SOLID  GEOMETRY 


452.    Theorem.  —  The  volume   of  a  sphere  is  equal  to  tht 
product  of  the  area  of  its  surface  and  one  third  of  the  radius. 


Hypothesis.  F=the  volume,  AS'=the  area  of  the  surface, 
and  R  =  the  radius  of  a  sphere. 

Conclusion.      V=  ^  MS. 

Proof.  1.  V,  S,  and  B  are  the  volume,  area  of  surface, 
and  radius,  respectively,  of  sphere.  Hyp. 

2.  Let  A0^  be  the  diameter  of  the  semicircle  by  revolv- 
ing which  the  sphere  is  traced.  Inscribe  in  the  semicircle 
one  half  of  a  regular  polygon  of  an  even  number  of  sides, 
ABOD  •...  Produce  any  side,  say  BQ,  to  meet  ^(7  produced 
at  some  point  H.  Draw  OB  and  00,  Draw  OMA.BC. 
Draw  BPl.Aa  and  OQl-Aa. 

3.  Vol.  traced  by  0  Off 

=  vol.  traced  by  OOQ-^  vol.  traced  by  OQH 
=  j7rx  Ce'xO^-f  Ittx  C^xG^  §414 
=  I  TT  X  (7$^  X  (  0^  -h  QH)  Factoring 

=  ^7r  xOQxOQx  Off.  Ax.  XII 

4.  But  it  may  be  proved  that  A  OMff'^A  OQff.  (Proof 
left  to  student.) 

-^  =  -^.  Def.  Sim.  A 

OM     OH 


5. 


THE   SPHERE  421 

6.  .-.  CQ  X  0E=  OMx  OH.  Ax.  IV 

7.  .-.  vol.  traced  hy  0011=  ^ir  x  CQ  x  OMx  OR. 

Ax.  XII 

8.  But  TT  X  OQ  X  0H=  area  of  surface  traced  by   OH. 

§410 

9.  .-.  vol.  traced  by  OOH^  \  OMx  area  traced  by  OH. 

Ax.  XII 

10.  Similarly,  vol.  traced  by  OBH=^  OiJf  xarea  traced 
hy  BH. 

11.  .-.  vol.  traced  by  OBO 

=  I  OMx  area  traced  by  OH—  ^  OMx  area  traced  by  BH 
=  ^OMx  area  traced  by  BO.  Factoring,  Ax.  XII 

12.  Similarly,  vol.  traced  by  OAB  =  ^  OMxarea,  traced 
by  AB,  vol.  traced  by  0OD  =  \  Oitfx  area  traced  by  CD, 
etc. 

13.  .*.  if  V  is  volume  traced  by  half  polygon  ABOD  •••,  and 
8  is  the  area  traced  by  its  perimeter, 

v  =  \  OMx  (area  traced  by  AB  +  area  traced  by  5(7+  etc.) 
=  \  OMx  8.  Ax.  II,  Factoring,  Ax.  XII 

14.  If  the  number  of  sides  of  the  polygon  is  indefinitely 
increased,  v=V,  8  =  S,  and  0M=  E.  §  447,  §  274 

15.  .'.iOMx8  =  i RS.  §  275  (2),  §  408  (2) 

16.  .-.  V=IRS.  §275  (1) 

453.  Corollary  i.  —  The  volume  of  a  8phere  equah  ^  7rM% 
or  J  ttZ)^,  where  R  18  the  radiu8  and  D  the  diameter. 

The  proof  is  left  to  the  student. 

454.  Corollary  2.  —  The  volumes  of  any  two  spheres  are  to 
each  other  as  the  cubes  of  their  radii.,  or  as  the  cubes  of  their 
diameters. 

The  proof  is  left  to  the  student. 


422 


SOLID  GEOMETRY 


455.  A  spherical  sector.  —  When  a  sphere  is  traced  by  re- 
volving a  semicircle  about  its  diameter  as  an  axis,  that  por- 
tion of  the  sphere  which  is  traced  by  a  sector  of  the  circle  is 
called  a  spherical  sector. 


Thus,  if  the  sector  A  OB  of  a  circle  is  revolved  about  the  diameter  EF 
as  an  axis,  it  traces  the  spherical  sector  AB-O-CD. 

The  zone  traced  by  the  arc  of  the  sector  of  a  circle  is  called 
the  base  of  the  spherical  sector. 

456.  Corollary  3.  —  The  volume  of  a  spherical  sector  is  equal 
to  the  product  of  the  area  of  its  base  and  one  third  of  the  radius 
of  the  sphere.     (Apply  §  452.) 

457.  A  spherical  segment.  —  A  spherical  segment  is  the  por- 
tion of  a  sphere  included  between  two  parallel  planes.  The 
sections  of  the  sphere  made  by  the  planes  are  the  bases,  and 
the  perpendicular  distance  between  the  planes  is  the  altitude. 


When  one  of  the  bounding  planes  is  tangent  to  the  sphere, 
the  segment  is  called  a  spherical  segment  of  one  base. 


THE  SPHERE 


423 


458.  Theorem.  —  The  volume  of  a  spherical  segment  of  which 
the  altitude  is  h  and  of  which  the  radii  of  the  bases  are  a  and 
b,  respectively,  equals  |  'Trh(a?'  +  ^^)  +  J  tt^^. 


Hypothesis, 
the  altitude 
Conclnsion. 
Proof.    1. 


The   volume  of    a   spherical    segment   is    FJ 
is  A,  and  the  radii  of   its  bases  are  a  and  b, 
V=  i  irh  (a2  +  62)  -h  1  TT^s. 
Vol.  sph.  seg.  =  V,  alt.  =  A,  etc.  Hyp. 

2.  Let  the  spherical  segment  be  traced  by  revolving 
DNMO,  one  half  of  a  segment  of  a  circle,  about  diameter 
AB  as  axis.  Then  OB  =  R,  MN=  h,  CM=  a,  and  I)N=  b. 
Draw  00  and  OD.     Let  ON  =  x  and  0M=  y, 

3.  Then  r=  vol.  traced  by  0(7Z>4-vol.  traced  by  OBliJ 
-  vol.  traced  h^OCM=\  irB^h  4-  J  irly^x  -  \  ira^y.  §  456,  §  414 

4.  But  A  =  a:  -  ?/,  62  =  i22  _  2^^  ^2  ^  722  _  ^2.  §  igg 

5.  .-.    r=  J  TT  S2  i^(2:  -  ?/)  4-  (i^2  _  3,2)^  _  (i22  _  ^2)^1 

=   ^7r(2:-^)    53i22_(^2^.^^4.^2){ 

=;|  ttAS  3  i22  _  (a^2  +  a:i/  +  «^2)  j .       Ax.  XII,  Fact. 
But  x^-2xy-\-y^=¥.  Ax.  VI 

Subtracting  this  from  identity  3  a:^  _|.  3  y2  __  3  aj2  ^  3  ^2, 

2  a;2  4.  2  2;?/  +  2  ^2  ^  3  3,2  _^  3  ^2  _  ^2. 
.-.  ic2  +  2:1/  +  ?/2  =  1(2^2  4-  2/2)  _  1  ^2  Ax.  V 

=  |(i?2  -  62  4-  i22  -  a2)  _  J  ^2  Ax.  XII 
=  3  i22  _  J  (^2  _^  52)  _  1  ^2. 
.-.  F=  J  rf  (a2  4-  52)  4. 1  ^^3.  Ax.  XII 


6. 

7. 

8. 


9. 


424  SOLID   GEOMETRY 

EXERCISES 

1.  Find  the  volume  of  a  sphere  whose  diameter  is  16  in. 

2.  Find  the  volume  of  a  spherical  sector  if  its  altitude  is  3  in.  and 
the  radius  of  the  sphere  is  8  in. 

3.  Find  the  volume  of  a  spherical  segment  if  its  altitude  is  4  in.  and 
the  radii  of  its  bases  are  12  in.  and  16  in.,  respectively. 

4.  If  V  is  the  volume  of  a  spherical  sector  and  H  is  the  altitude  of  its 
base,  and  R  is  the  radius  of  the  sphere,  prove  that  F  =  f  tt  R^H. 

5.  Prove  that  the  volume  of  a  spherical  segment  of  one  base  equals 
7rh^(R  —  i  A),  where  h  is  the  altitude  and  R  is  the  radius  of  the  sphere. 

Suggestion.  —  In  the  formula  of  §  458,  let  &  =  0,  and  apply  §  200. 

6.  Find  the  ratio  of  the  volume  of  a  sphere  to  that  of  a  circumscribed 
cube. 

7.  Find  the  ratio  of  the  volume  of  a  sphere  to  that  of  an  inscribed 
cube. 

8.  Prove  that  the  volume  of  any  sphere  is  equal  to  two  thirds  of  the 
volume  of  the  circumscribed  right  circular  cylinder. 

9.  A  sphere  whose  diameter  is  D  is  transformed  into  a  cylinder  of 
revolution  having  the  same  diameter.     Find  the  height  of  the  cylinder. 

10.  Show  that  the  formula  for  computing  the  volume  of  a  sphere 
may  be  obtained  from  the  formula  for  computing  the  volume  of  a 
spherical  sector. 

11.  Show  that  the  formula  for  computing  the  volume  of  a  sphere 
may  be  obtained  from  the  formula  for  computing  the  volume  of  a 
spherical  segment. 

12.  Prove  that  if  a  sphere  and  a  cube  have  equal  areas,  the  sphere 
has  the  greater  volume. 

Note.  —  It  is  interesting  to  note  that  of  all  solids  having  equal  areas,  the 
sphere  has  the  greatest  volume. 

13.  Two  spheres  of  lead,  of  radii  2  in.  and  4  in.,  respectively,  are 
melted  and  recast  into  a  solid  cylinder  of  revolution  whose  altitude  is 
6  in.     Show  that  the  total  surface  is  unchanged  in  amount. 

14.  A  cubic  foot  of  ivory  weighs  114  lb.  What  is  the  weight  of  an 
ivory  billiard  ball  2  in.  in  diameter? 


THE   SPHERE  425 

15.  A  hollow  spherical  steel  shell  is  1  in.  thick,  and  its  outer  diameter 
is  10  in.     Allowing  490  lb.  to  a  cubic  foot,  find  its  weight. 

16.  A  washbasin  is  in  the  form  of  a  spherical  segment  of  one  base. 
Its  depth  is  7^  in.  and  the  distance  across  the  top  is  16  in.  Find  how 
many  gallons  of  water  it  will  hold,  allowing  7^  gal.  to  a  cubic  foot. 

17.  Through  the  center  of  a  10-in.  sphere  a  hole  was  bored,  taking 
off  one  fourth  of  the  surface  of  the  sphere  from  each  end.  Find  the 
volume  of  the  ring  left. 

18.  A  boiler  is  made  in  the  form  of  a  4-ft.  cylinder  2  ft.  in  diameter, 
with  hemispherical  ends.     How  many  gallons  will  it  hold? 

19.  A  sphere  of  lead  5  inches  in  diameter  is  formed  into  a  tube  2^  ft. 
long  whose  internal  diameter  is  1  in.     Find  the  thickness  of  the  tube. 

20.  A  hemispherical  boiler  whose  diameter  is  4  ft.  contains  water  to 
a  depth  of  20  in.     How  many  gallons  of  water  does  it  contain? 

21.  Among  the  rules  of  thumb  used  by  men  in  practical  work  is  the 
following:  The  weight  of  a  cast-iron  ball  (in  pounds)  equals  the  cube 
of  the  diameter  times  .1377.  Allowing  450  lb.  to  a  cubic  foot  of  cast 
iron,  test  the  accuracy  of  this  rule. 

22.  Which  is  the  better  bargain,  oranges  3  in.  in  diameter  at  30^  a 
dozen,  or  oranges  3^  in.  in  diameter  at  40  ^  a  dozen  ? 

23.  The  figure  represents  a  cross  section  of  a 
ball  safety  valve.  The  pressure  of  the  steam  from 
below  lifts  the  ball,  and  allows  the  steam  to  escape. 
The  diameter  of  the  pipe  below  the  ball  is  |  in.,  and 
the  diameter  of  the  ball  is  1  in.  If  the  ball  weighs 
.28  lb.  to  the  cubic  inch,  what  is  the  pressure  in 
pounds  per  square  inch  of  the  steam  in  the  pipe 
below  the  ball  when  it  lifts  it  and  escapes  ? 

24.  A  haystack  is  approximately  in  the  form  of 
a  cylinder  16  ft.  in  diameter  and  8  ft.  high,  surmounted  by  a  hemi- 
sphere.    Allowing  512  cu.  ft.  to  the  ton,  find  the  weight  of  the  stack. 

25.  A  building  lot  which  is  being  leveled  off  has  a  mound  on  it 
approximately  in  the  form  of  a  spherical  segment  of  one  base.  It  is 
40  ft.  wide  at  the  base  and  10  ft.  deep  at  the  highest  point.  How  many 
cubic  yards  of  earth  must  be  removed  in  cutting  it  down  to  the  level  of 
the  surrounding  ground  ? 


426  SOLID  GEOMETRY 

26.  When  an  object  floats  in  water,  or  is  immersed  in  it,  the  object 
is  buoyed  up  by  the  water  with  a  force  equal  to  the  weight  of  the  water 
displaced  by  it. 

A  ball  floats  half  submerged  in  water.    Its  diameter  is  8  in.     Find  its 
weight.      (Water  weighs    62.5  lb.  per 
cubic  foot.) 

27.  A  float  in  a  water  tank  consists 
of  a  hollow  copper  ball  which  closes  the 
valve  that  lets  the  water  enter  the  tank, 
by  means  of  a  lever  to  which  it  is 
attached,  when  the  water  buoys  it  up 
to  a  certain  point.  The  valve  shuts  ofE  the  entering  water  when  the 
ball  is  submerged  to  three  fourths  of  its  depth.  If  the  diameter  of  the 
float  is  4  in.,  what  is  the  upward  pressure  of  the  water  against  it  ? 

FIGURES   ON   THE   SURFACE   OF   A   SPHERE 

459.  Spherical  angles.  —  A  spherical  angle  is  the  figure 
formed  by  two  arcs  of  great  circles  drawn  from  the  same 
point  on  the  surface  of  a  sphere.  The 
point  from  which  the  arcs  are  drawn  is  the 
vertex  of  the  angle,  and  the  arcs  are  the 
sides  of  the  angle. 

A  spherical  angle  is  measured  by  the 
plane  angle  which  is  formed  by  the  tan- 
gents to  its  sides  at  the  vertex. 

Thus,  spherical  angle  ABC  is  measured  by  Z  DBE,  where  BD  and 
BE  are  tangents  to  BA  and  BC  respectively. 

A  spherical  angle  of  which  the  tangents  to  the  sides  at 
the  vertex  form  a  right  angle  is  called  a  right  spherical  angle. 

460.  Perpendicular  arcs.  —  The  sides  of  a  right  spherical 
angle  are  perpendicular  arcs. 

Thus,  if  ^  and  ^C  in  the  figure  of  §  469  form  a  right  spherical  angle, 
they  are  called  perpendicular  arcs. 


FIGURES  ON   THE   SURFACE   OF  A  SPHERE        427 

461.  Theorem.  — A  spherical  angle  has  the  same  measure  as 
the  arc  of  the  great  circle  having  the  vertex  of  the  angle  as  pole 
and  included  between  the  sides  of  the  angle^  produced  if 
necessary. 

A 


Hypothesis.  Z.BAO  is  a  spherical  angle  on  the  sphere 
with  center  0\  BCis  the  arc  of  a  great  circle  having  A  as 
its  pole. 

Conclusion.     ^  BAOha^  the  same  measure  as  BO.    , 
Proof.     1.    Z  BAG  is  spherical  angle;  BO  is  arc  of  great 
circle  having  A  as  pole.  Hyp. 

2.  Draw  radii  OA,  OB,  00,     Draw  AD  tangent  to  AB 
and  AU  tangent  to  AO. 

3.  OA  ±  plane  BOO.  §  436 

4.  .-.  OB±OA  and  001.  OA,  §  302 

5.  AB  ±  OA  and  AH  ±  OA.  §  164 

6.  .-.  AB  II  OB  and  AU  W  00.  §  41 

7.  .'.Z.DAE=ZBOO.  §  317 

8.  But  Z  BAO  has  the  same  measure  as  Z.  DAE.      §  459 

9.  And  BO  has  the  same  measure  d^s  Z  BOO.  §  173 
10.    .*.  Z  BAOhdi^  the  same  measure  as  BO,               Ax.  I 

462.    Corollary  l.  —  A  spherical  angle  has  the  same  measure 
as  the  diedral  angle  formed  by  the  planes  of  its  sides. 
The  proof  is  left  to  the  student. 


428  SOLID   GEOMETRY 

463.  Corollary  2.  —  An  are  of  a  great  eirele  drawn  through 
the  pole  of  another  great  circle  is  perpendicular  to  that  circle. 

The  proof  is  left  to  the  student. 

464.  Spherical  polygons.  —  A  spherical  polygon  is  the  figure 
formed  by  three  or  more  arcs  of  great  circles  which  com- 
pletely inclose  a  portion  of  the  surface  of  a  sphere. 

The  arcs  are  the  sides,  the  spherical  angles  formed  by  the 
sides  are  the  angles,  and  the  vertices  of  the  angles  are  the 
vertices  of  the  polygon. 

A  diagonal  of  a  spherical  polygon  is  the 
great  circle  arc  which  joins  any  two  non- 
adjacent  vertices. 

A  spherical  polygon  each  of  whose  angles 
is  less  than  180°  is  convex.  Only  convex 
spherical  polygons  are  considered  here. 

A  spherical  triangle  is  a  spherical  polygon  of  three  sides. 

The  terms  right  triangle,  isosceles,  equilateral,  etc.,  are 
applied  to  spherical  triangles  with  the  same  meanings  as 
when  applied  to  plane  triangles. 

465.  Relation  of  spherical  polygons  to  polyedral  angles.  —  The 

planes  of  the  sides  of  a  spherical  polygon  form  a  polyedral 
angle  whose  vertex  is  the  center  of  the  sphere. 

Thus,  in  the  figure  of  §  464,  the  planes  of  the  sides  of  polygon  ABCD 
form  the  polyedral  angle  0-ABCD. 

It  is  evident  from  the  preceding  sections  that: 

(1)  Each  face  angle  of  the  polyedral  angle  has  the  same 
measure  as  the  corresponding  side  of  the  spherical  polygon. 

(2)  Uach  diedral  angle  of  the  polyedral  angle  has  the  same 
measure  as  the  corresponding  angle  of  the  spherical  polygon. 

Hence^  from  any  property  of  polyedral  angles  an  analogous 
property  of  spherical  polygons  may   he   inferred. 


FIGURES  ON   THE   SURFACE   OF  A  SPHERE        429 

466.    Theorem.  — Any  side  of  a  spherical  triangle  is  less  than 
the  sum  of  the  other  two  sides. 


The  proof  is  left  to  the  student.     See  §  342  and  §  425  (1). 
Write  the  proof  in  full. 

467.    Theorem.  —  The  sum  of  the  sides  of  any  spherical  poly- 
gon is  less  than  360°. 


The  proof  is  left  to  the  student.     Write  the  proof  in  full. 

EXERCISES 

1.  The  perimeter  of  any  spherical  polygon  is  less  than  a  great  circle. 

2.  Any  side  of  a  spherical  polygon  is  less  than  the  sum  of  the  others. 

3.  No  side  of  a  convex  spherical  polygon  is  as  long  as  a  semicircle. 

4.  If  two  sides  of  a  spherical  triangle  are  quadrants,  the  angles  oppo- 
site them  are  right  angles. 

5.  If  two  sides  of  a  spherical  triangle  are  80°  and  90°,  respectively, 
between  what  two  values  does  the  third  side  lie  ? 


430 


SOLID  GEOMETRY 


468.  Polar  triangles.  —  If  A  ABCis  any  spherical  triangle, 
and  if  the  three  great  circles  are  drawn  of  which  A^  B^  and 
(7,  respectively,  are   poles,  these   circles 

evidently  form  eight  new  spherical  tri- 
angles. Of  these,  one  triangle  BEF  is 
called  the  polar  triangle  of  A  ABC.  A  is 
the  pole  of  EF^  B  is  the  pole  of  DF^  and 
O  is  the  pole  of  BE,  Of  the  eight  tri- 
angles, A  BEF  is  so  chosen  that  0  and  F 
are  on  the  same  side  of  AB^  A  and  B  on 
the  same  side  of  BO^  and  B  and  E  on  the  same  side  of  AC. 

Note. — The  properties  of  polar  triangles  were  discovered  by  Girard,  a 
Dutch  mathematician,  about  1626  a.d.  They  were  discovered  independently 
about  the  same  time  by  Snell,  a  prodigy,  who  at  the  age  of  twelve  was  familiar 
with  the  standard  works  on  higher  mathematics  of  that  time. 

469.  Theorem.  —  If  one  spherical  triangle  is  the  polar  of  a 
second,  the  second  spherical  triangle  is  the  polar  of  the  first. 


Hypothesis.     A  BEF  is  the  polar  of  A  ABC. 
Conclusion.     A  ABC  is  the  polar  of  A  BEF. 

Proof.     1.    A  BEF  is  the  polar  of  A  ABC.  Hyp. 

2.  Draw  BB  and  CB  of  great  circles. 

3.  Since  B  is  the  pole  of  BF,  BB  is  a  quadrant.  §  440 

4.  Since  C  is  the  pole  of  BE,  CB  is  a  quadrant.  §  440 

5.  .-.  B  is  the  pole  of  ^.  §  441 

6.  Similarly,  E  is  the  pole  of  AC  and  F  the  pole  of  AB. 

7.  .-.  A  ABCis  the  polar  of  A  BEF.  §  468 


FIGURES   ON  THE   SURFACE   OF  A  SPHERE        431 

470.  Theorem.  —  In  two  polar  spherical  triangles^  each  angle 
of  one  has  the  same  measure  as  the  supplement  of  the  side  of  the 
other  of  which  its  vertex  is  the  pole. 


Hjrpothesis.  A  ABC  and  aDEF  are  polar  spherical  tri- 
angles. 

Conclusion.  Z  C  has  the  same  measure  as  the  supplement 
of  DE^  Z  F  has  the  same  measure  as  the  supplement  of 
ABy  etc. 

Proof.     1.    A  AjBC  and  A  BEF  are  polar  triangles.     Hyp. 

2.  Prolong  CA  and  (S,  if  necessary,  to  meet  BE  at  M 
and  iV,  respectively. 

3.  Then  Z  (7  has  the  same  measure  as  MN,  §  461 

4.  ^=90°andMfe=90°.  §440 

5.  .-.  BN^-ME=  180°.  Ax.  II 

6.  ByxtBN+ME=MN+BE.  ^  Ax.  XII 

7.  .'.  MN+  BE^  180°,  i.e.  ilEV'and  BE  are  supp.     Ax.  I 

8.  .'.  Z  (7  has  the  same  measure  as  the  supplement  of  BE. 

Ax.  XII 

9.  Similarly,  Z  F  has  the  same  measure  as  the  supplement 


of  AB.,  etc. 

471.    Corollary  l.  —  If  a  spherical  triangle  is  equilateral,  its 
polar  triangle  is  equiangular  ;  and  conversely. 

The  proof  is  left  to  the  student. 


432  SOLID  GEOMETRY 

472.  Definitions.  —  Two  mutually  equiangular  triangles  (See 
§  127)  are  triangles  such  that  to  each  angle  of  one  there 
corresponds  an  equal  angle  of  the  other.  Two  mutually 
equilateral  triangles  are  triangles  such  that  to  each  side  of 
one  there  corresponds  an  equal  side  of  the  other. 

473.  Corollary  2. — If  two  spherical  triangles  on  the  same 
sphere^  or  on  equal  spheres^  are  mutually  equiangular^  their 
polar  triangles  are  mutually  equilateral ;  and  conversely. 

The  proof  is  left  to  the  student. 

474.  Theorem.  —  The  sum  of  the  angles  of  a  spherical  tri- 
angle is  greater  than  180°  and  less  than  540°. 


Hypothesis.     A  ABO  i^  any  spherical  triangle. 
Conclusion.     Z^  +  Z5-fZ(7>  180°  and  <  540°. 
Proof.     1.    Let  d^  e,  and  /  be   the   sides  of   the   polar   of 
AABQoi  which  ^,  B,  and  (7,  respectively,  are  the  poles. 

2.  Then  Z  A  has  the  same  measure  as  180°  —  d, 

Z.  B  has  the  same  measure  as  180°—  e, 

Z  (7  has  the  same  measure  as  180°  — /.  §  470 

3.  .'./.A-^Z.B-\-/.0  has  the  same  measure  as 

540° -((^  +  g+/).  Ax.  Ill 

4.  Butc^  +  e+/<360°and  >0°.  §467 

5.  .-.  540°-((^4-e-f/)>  180°  and  <  540°.      Subtraction 

6.  .-.    Z^-|-Z^  +  Z(7>180°and  <540°.  Ax.  XII 


FIGURES  ON  THE   SURFACE   OF  A  SPHERE        433 

475.  Corollaxy.  —  A  spherical  triangle  may  have  one^  two^  or 
three  right  angles.  And  it  may  have  one^  two^  or  three  obtuse 
angles. 

The  proof  is  left  to  the  student. 

476.  Definitions.  —  A  spherical  triangle  having  two  right 
angles  is  a  birectangular  spherical  triangle.  One  having 
three  right  angles  is  a  trirectangpilar  spherical  triangle. 

The  difference  between  the  sura  of  the  angles  of  any  spheri- 
cal triangle  and  180°  is  the  spherical  excess  of  the  triangle. 

EXERCISES 

1.  The  angles  of  a  spherical  triangle  are  80°,  65°,  and  135°.  Find  the 
sides  of  its  polar  triangle. 

2.  The  sides  of  a  spherical  triangle  are  120°,  84°,  and  66°,  respectively. 
How  many  degrees  are  there  in  each  angle  of  its  polar  triangle  ? 

3.  A  spherical  triangle  has  two  of  its  sides  quadrants  and  the  third 
side  equal  to  60°.     Determine  the  angles  of  its  polar  triangle. 

4.  If  two  sides  of  a  spherical  triangle  are  quadrants,  the  triangle  is 
birectangular. 

5.  If  a  spherical  triangle  is  birectangular,  the  sides  opposite  the  right 
angles  are  quadrants. 

6.  Three  planes  passed  through  the  center  of  a  sphere,  each  perpen- 
dicular to  the  other  two,  form  on  the  spherical  surface  eight  trirectangular 
triangles. 

7.  Find  the  sides  of  a  spherical  triangle  if  the  angles  of  its  polar 
triangle  are  96°  37'  36",  87°  17'  57",  72°  46'  32". 

8.  What  is  the  polar  triangle  of  a  spherical  triangle  all  of  whose 
sides  are  quadrants? 

9.  In  any  birectangular  spherical  triangle  the  side  opposite  the  angle 
that  is  not  a  right  angle  has  the  same  measure  as  that  angle. 

,      10.  In  any  trirectangular  spherical  triangle  each  of  the  sides  is  a 
quadrant. 

11.  The  angles  of  a  spherical  triangle  are  80°,  90°,  and  120°.  What 
is  the  spherical  excess  ? 


434  SOLID  GEOMETRY 

12.  Any  trirectangular  spherical  triangle  coincides  with  its  polai 
triangle. 

13.  Any  exterior  angle  of  a  spherical  triangle  is  less  than  the  sum  of 
the  two  opposite  interior  angles. 

14.  The  arcs  of  the  great  circles  which  bisect  the  angles  of  a  spherical 
triangle  are  concurrent. 

15.  Any  arc  of  a  great  circle  which  is  perpendicular  to  a  second  great 
circle  passes  through  the  pole  of  the  latter. 

477.  Symmetrical  spherical  triangles.  —  Two  triangles  on  the 
same  sphere,  or  on  equal  spheres,  are  symmetrical  when  to  each 
part  of  one  there  corresponds  an  equal  part  of  the  other,  but  the 
equal  parts  are  arranged  in  reverse  order. 

It  is  easily  seen  that  if  three  planes 
pass  through  the  center  of  a  sphere  and 
do  not  intersect  in  one  straight  line,  they 
cut  the  surface  of  the  sphere  in  a  pair 
of  symmetrical  spherical  triangles,  as 
A  ABQ  and  A  DEF  in  the  figure.  Also 
it  may  be  assumed  that  two  symmetrical 
triangles  of  a  sphere  may  be  moved  into  such  positions  as  are 
occupied  by  A  ABO  and  A  DEF. 

It  is  evident  that,  because  of  the  curvature  of  the  spherical 
surface,  in  general   two  symmetrical 
spherical  triangles  cannot  be  made  to 
coincide,  and  hence  are  not  congruent. 
See  §  344. 

If,  however,  two  symmetrical  spheri-      j} ]g     £) 

cal  triangles  are  isosceles^  as  A  ABC 

and  A  DEF  in  the  figure,  it  is  possible  to  make  them  coincide 

throughout.     Hence  it  is  inferred  that : 

If  two  symmetrical  spherical  triangles  are  isosceles,  they  are 
congruent. 


C  F 


FIGURES  ON  THE  SURFACE   OF  A  SPHERE        435 

478.   Theorem.  —  Two  symmetrical  spherical  triangles  have 
equal  areas. 


Hjrpothesis.  A  ABC  and  ADEF  are  symmetrical  spheri- 
cal triangles. 

Conc^asion.    AABC^aBEF, 

Proof.  1.  A  ABO  and  A  BEF  are  symmetrical  spherical 
triangles.  Hyp. 

2.  .'.A  ABC  and  A  BEF  may  be  placed,  as  in  the  figure, 
so  that  AB,  BE,  and  CF  are  diameters.  §  477 

3.  Let  P  be  the  pole  of  the  circle  determined  by  points 
A,  jB,  (7,  and  let  P^  be  a  diameter.  Draw  great  circle  arcs 
AP,  BP,  CP,  BQ,  EQ,  FQ. 

4.  Then  AP=^BP=CP.  §  438 

5.  AP  =  BQ,BP  =  fQ,6P=f^.  §20(V),  §173 

6.  .'.BQ=EQ  =  FQ.  Ax.I 

7.  .-.  the  symmetrical  A  APB  and  BQE  are  isosceles. 

§464 

8.  .'.AAPB^aBQE.  §477 

9.  Similarly,  A  BPC  ^  A  EQF,  A  CPA  ^  A  FQB. 

10.    .'.AABC^A  BEF.  Ax.  II 

EXERCISE 

Draw  a  figure  for  §  478  in  which  the  pole  P  falls  outside  of  A  ABC 
and  Q  falls  outside  of  A  DEF,  and  give  the  proof  of  the  theorem. 

Suggestion.  —  Each  of  the  given  triangles  will  be  equivalent  to  the 
sum  of  two  isosceles  triangles  minus  a  third  one. 


436  SOLID  GEOMETRY 

479.  Theorem. — If  two  spherical  triangles  on  the  same 
sphere  or  equal  spheres  have  two  sides  and  the  included  angle 
of  one  equal  respectively  to  two  sides  and  the  included  angle 
of  the  other,  they  are  either  congruent  or  symmetrical. 


Hypothesis.  A  ABO  and  A  DUF  are  spherical  triangles 
on  the  same  sphere  or  equal  spheres  ;  AB  =  BE,  A0=  BF, 
ZA^ZB. 

Conclusion.  A  ABO  and  A  BFJF  are  either  congruent  or 
symmetrical. 

Suggestions.  If  the  equal  parts  of  the  triangles  occur  in 
the  same  order,  superpose  A  BFF  upon  A  ABO  so  that  Z  B 
coincides  with  Z.A,  and  prove  that  the  triangles  coincide 
throughout. 

If  the  equal  parts  occur  in  reverse  order,  construct  A  GrHK 
symmetrical  to  A  BEF.  Prove  that  A  ABO  ^.  A  GHK,  and 
hence  that  A  ABO  i^  symmetrical  to  A  BEF. 

Write  the  proof  in  full. 

480.  Theorem.  —  If  two  spherical  triangles  on  the  same 
sphere  or  equal  spheres  have  a  side  and  two  adjacent  angles  of 
one  equal  respectively  to  a  side  and  two  adjacent  angles  of  the 
other,  they  are  either  congruent  or  symmetrical. 

The  proof  is  left  to  the  student.  Proceed  as  in  the  proof  of 
§  479. 

Write  the  proof  in  full. 


FIGURES  ON  THE   SURFACE   OF  A  SPHERE        437 


481.  Theorem. — If  two  spherical  triangles  on  the  same 
sphere  or  equal  spheres  are  mutually  equilateral^  they  are  either 
congruent  or  symmetrical. 

C  F  F 


D'^" 


y^^^-^D 


Left  to  the  student.  Prove  diedral  angles  with  edges  A  0 
and  DP  equal,  and  that  Z  ^  =  Z  D.     Apply  §  479. 

482.  Theorem.  —  If  two  spherical  triangles  on  the  same 
sphere  or  equal  spheres  are  mutually  equiangular^  they  are 
either  congruent  or  symmetrical. 


K 


N 


N 


H 


M  '^ 


Hypothesis.  A  ABQ  2^vA  A  DEF  are  triangles  on  the  same 
sphere  or  equal  spheres ;  /.A  =  Z.D^Z.B  =  Z.I]^/.0=/.F. 

Conclusion.  A  ABO  and  A  DEF  are  either  congruent  or 
symmetrical. 

Suggestions.  Draw  A  GrRK ^nd  A  LMN"  polars  of  A  ABO 
and  A  DEF.  Prove  A  GrRKsmd  A  LMN  mutually  equilat- 
eral. Then  they  are  congruent  or  symmetrical.  Why  ? 
Hence  they  are  mutually  equiangular.  But  A  ABO  and 
A  DEF  are  polars  of  A  GrIIKsind  A  LMJY.  Now,  from  these 
facts,  by  similar  steps,  show  that  A  ABO  and  A  DEF  2iTe 
congruent  or  symmetrical. 


438  SOLID  GEOMETRY 

483.   Theorem. — In    any    isosceles    spherical    triangle    the 
angles  opposite  the  equal  sides  are  equal. 


Draw  a  great  circle  arc  from  (7  to  Z>,  the  middle  point  of  Ah, 
Compare  A  ^1>(7  and  A  i>J?(7  by  §  481. 
Write  the  proof  in  full. 

484.  Theorem.  —  If  two  angles  of  a  spherical  triangle  are 
equals  the  sides  opposite  these  angles  are  equals  and  the  tri- 
angle is  isosceles,- 


.-'% 


Hypothesis.     In  spherical  A  ABO^  Z  A  =  Z  B. 

Conclusion.    AO=BO. 

Proof.     1.    In  spherical  AABC,ZA  =  ZB.  Hyp. 

2.  Let  A  BUF  be  the  polar  of  A  ABO, 

3.  Then  JDF  and  JEF  have  the  same  measures  as  the  sup- 
plements oi  Z  B  and  Z  A^  respectively.  §  470 

4.  ,',]5f=EF,  Supp.  of  equals 

5.  .-.  ZD=ZE,  §483 

6.  But  A  A^CMs  the  polar  of  A  DJ^^.  §469 


FIGURES  ON  THE  SURFACE   OF  A  SPHERE        439 

7.  .*.  AO  and  ^(7  have  the  same  measures  as  the  supple- 
ments of  Z  ^  and  Z  i>,  respectively.  §  470 

8.  .-.  JX/=  Bu.  Supp.  of  equals 

EXERCISES 

1.  The  great  circle  arc  from  the  vertex  of  an  isosceles  spherical  tri- 
angle to  the  middle  point  of  the  base  bisects  the  vertical  angle,  is  per- 
pendicular to  the  base,  and  divides  the  triangle  into  two  triangles  with 
equal  areas. 

2.  Every  point  in  the  arc  of  a  great  circle  perpendicular  to  another 
arc  of  a  great  circle  at  its  middle  point  is  equidistant  from  the  ends  of 
that  arc. 

3.  Every  point  on  the  surface  of  a  sphere  that  is  equidistant  from  the 
ends  of  an  arc  of  a  great  circle  is  on  the  arc  perpendicular  to  that  arc  at 
its  middle  point. 

4.  What  is  the  locus  of  points  on  the  surface  of  a  sphere  that  are 
equidistant  from  the  ends  of  a  given  great  circle  arc  ? 

5.  If  an  arc  of  one  great  circle  is  perpendicular 
to  an  arc  of  another  great  circle  at  its  middle 
point,  any  point  without  the  former  is  unequally 
distant  from  the  ends  of  the  latter. 

6.  Any  point  on  the  bisector  of  a  spherical 
angle  is  equally  distant  from  the  sides  of  the  angle. 

Suggestion.  —  Let  BD  bisect  ZABC,  and  let 
MP  be  JL  AB.  Mark  ofE  jBQ  =  BP,  and  draw  MQ. 
Prove  that  MQ±BC  and  MP  =  MQ. 

7.  In  any  spherical  triangle,  if  two  angles  are  unequal,  the  opposite 
sides  are  unequal,  the  greater  side  being  opposite  the  greater  angle. 

Hypothesis.  —  In  spherical  A  ABC, 
ZBAOZCBA. 

Conclusion.  —  sd  >  AC. 

Suggestions.  —  Draw  AD  making  Z  BAD 
=  Z  CBA.  Prove  AD  =  BD.  Then  AD  + 
DC' >  AC,  etc. 

8.  State  and  prove  the  converse  of  Exercise  7. 


440  SOLID  GEOMETRY 

485.  Area  of  a  trirectangular  triangle.  —  If  three  planes  are 
passed  through  the  center  of  a  sphere,  each  perpendicular  to 
the  other  two  planes,  they  form  on  the  surface 
of  the  sphere  eight  trirectangular  triangles^ 
which  are  equal  in  area  by  §  482.     Hence: 

The  area  of  a  trirectangular  triangle  is 
equal  to  one  eighth  of  the  area  of  the  surface 
of  the  sphere. 

Such  a  triangle  is  sometimes  used  as  a  unit  of  measure  of 
spherical  surfaces. 

486.  A  spherical  degree.  —  One  of  the  720  equal  parts  of 
the  surface  of  a  sphere  is  called  a  spherical  degree. 

A  spherical  degree  is  used  as  a  unit  of  measure  of  spherical 
surfaces. 

487.  A  lime.  —  A  lune  is  that  figure  on  the  surface  of  a 
sphere  which  is  formed  by  two  great  semicircles  whose  end 
points  coincide.     A  lune,  then,  has  two  equal  angles. 

Thus,  ABCD  is  a  lune  whose  equal  angles  are  Z  A  and  /.  C. 

It  is  evident  that  lunes  on  the  surface  of 
the  same  sphere  which  have  equal  angles  may 
be  made  to  coincide,  and  hence  are  congru- 
ent.    Consequently,  it  is  inferred  that: 

The  area  of  a  lune  whose  angle  is  n  degrees 

is  — -  of  the  area  of  the  surface  of  the  sphere. 

488.  Corollary.  —  The  area  of  a  lune.,  in  spherical  degrees^  is 
equal  to  twice  the  number  of  degrees  in  the  angle  of  the  lune. 

For,  by  §  486  and  §  487,  the  area  of  a  lune  whose  angle  is 

n  degrees,  is  equal  to  -^  x  720,  or  2  w,  spherical  degrees. 


FIGURES  ON  THE  SURFACE   OF  A  SPHERE        441 

489.    Theorem.  —  The  area  of  a  spherical  triangle^  in  spheri- 
cal degrees^  is  equal  to  the  spherical  excess  of  the  triangle. 


Hypothesis.     Spherical  excess  of  A  ABC  —  e. 
Conclusion.     The  number  of  sph.  deg.  in  A  ABQ  =  e. 
Proof.     1.    Spherical  excess  of  A  ABQ  =  e.  Hyp. 

2.  Produce  each  side  of  A  ABO  to  form  the  complete 
circle,  and  let  AF^  BU,  and  OB  be  the  diameters  in  which 
their  planes  intersect. 

3.  The  vertical  angrles  at  0  are  equal.  §  20,  V 

4.  .-.  Bd=  DE.  BF  =  AF,  OF^  lb.  §  173 

5.  .'.  A  ABE  and  A  BOF  are  either  congruent  or  sym- 
metrical. §  481 

6.  .-.  A  ABE  and  A  BOF  are  equal.       Def.  cong.,  §  478 

7.  ^o\Y\unQABFO=AABO+ABOF, 

lune  BOEA  =  A  ABO+  A  AOE,  Ax.  X 

luae  OBBA  =  A  ABO+  A  ABD. 

8.  .'.ABFO-\-BOEA+OBBA  = 

2AAB0+  (A  ABO-^  A  BOF-}-  aAOE-\-  A  ABB).    Ax.  II 

9.  Bnt  ABFO+BOEA+  0BBA  =  2iA-\-B+  0)  sph, 
deg.  §  488 

10.  And  A  J.5(7+A^(7^+A^(7^+A^5i)=a  hemi- 
spherical surface,  or  360  sph.  deg.  §  486 

11.  .'.2(A-\-B+  0)  sph.  deg.  =  2aABO-{-SQ0  sph.  deg. 

Ax.  XII 

12.  .-.  A^^(7=^  +  5+6'-180sph.  deg.  =e.    Solving 


442  SOLID  GEOMETRY 

490.  Spherical  excess  of  a  spherical  polygon.  —  The  sphericav 
excess  of  a  spherical  polygon  is  the  difference  between  the  sum 
of  its  angles  and  the  sum  of  the  angles  of  a  plane  polygon 
of  the  same  number  of  sides. 

491.  Theorem.  —  The  area  of  a  spherical  polygon^  in  spheri- 
cal degrees^  is  equal  to  the  spherical  excess  of  the  polygon. 

The  proof  is  left  to  the  student.  Draw  diagonals  from 
any  vertex,  dividing  the  polygon  into  triangles.  How  many 
triangles  are  there  ? 

Write  the  proof  in  fall. 

EXERCISES 

1.  Find  the  area  of  a  trirectangiilar  triangle  on  a  sphere  whose 
diameter  is  28  in. 

2.  Find  the  area  of  a  lune  whose  angle  is  30°  on  a  sphere  whose 
diameter  is  20  in. 

3.  What  part  of  the  whole  sphere  is  a  triangle  whose  angles  are  70°, 
110°,  and  120°  ? 

4.  What  is  the  spherical  excess  of  a  triangle  whose   angles  are 
84°  27'  20",  96°  42'  36",  and  116°  12'  24"  ? 

5.  What  is  the  area  of  a  triangle  whose  angles  are  80°,  92°,  and  112°, 
on  a  sphere  whose  diameter  is  36  in.  ? 

6.  What  is  the  spherical  excess  of  a  hexagon  whose  angles  are  84°, 
167°,  140°,  106°,  98°,  and  157°? 

7.  Find  the  area  of  a  spherical  pentagon  whose  angles  are  120°,  140°, 
155°,  158°,  and  163°,  on  a  sphere  whose  diameter  is  10  in.  . 

8.  Find  the  area  of  that  part  of  the  earth's  surface  lying  between  the 
75th  and  90th  meridians.     Call  the  radius  4000  mi. 

9.  The  areas  of  two  lunes  on  the  same  sphere  or  on  equal  spheres 
are  to  each  other  as  the  angles  of  the  lunes. 

10.  The  areas  of  lunes  having  equal  angles,  but  situated  on  unequal 
spheres,  are  to  each  other  as  the  squares  of  the  radii  of  the  spheres  on 
which  they  are  situated. 


FIGURES  ON  THE  SURFACE   OF  A  SPHERE        443 

MISCELLANEOUS  EXERCISES 

1.  The  radius  of  a  sphere  is  12  in.  Find  the  area  of  a  section  made 
by  a  plane  8  in.  from  the  center. 

2.  Find  the  locus  of  all  points  in  space  which  are  equidistant  from 
two  given  points  and  at  a  given  distance  d  from  a  third  given  point. 

3.  Any  two  vertical  spherical  angles  are  equal. 

4.  A  right  circular  cylinder  whose  altitude  is  8  in.  is  inscribed  in  a 
sphere  whose  radius  is  6  in.     Find  the  volume  of  the  cylinder. 

5.  Any  side  of  a  spherical  polygon  is  less  than  180°. 

6.  If  two  adjacent  sides  of  a  spherical  quadrilateral  are  greater, 
respectively,  than  the  other  two  sides,  the  angle  included  by  the  two 
shorter  sides  is  greater  than  the  angle  included  by  the  two  greater  sides, 

7.  The  radii  of  two  concentric  spheres  are  8  in.  and  12  in.,  respec- 
tively. A  plane  is  tangent  to  the  inner  sphere.  Find  the  area  of  the 
section  of  the  outer  sphere  made  by  the  plane.  Find  the  area  of  the 
smaller  zone  of  the  outer  sphere  cut  o£E  by  the  plane. 

8.  A  sphere  is  inscribed  in  a  right  circular  cylin- 
der, touching  the  lateral  surface  of  the  cylinder  all  the 
way  around  and  tangent  to  both  bases.  Two  cones 
have  the  bases  of  the  cylinder  as  their  bases  and  the 
center  of  the  sphere  as  their  common  vertex.  Any 
plane  is  passed  through  the  figure  parallel  to  the  bases 
of  the  cylinder.  Prove  that  the  ring  between  the  sec- 
tions of  the  cylinder  and  cone  is  equal  to  the  section 
of  the  sphere. 

9.  If  two  circles  on  a  sphere  have  the  same  poles,  their  planes  are 
parallel. 

10.  The  line  of  centers  of  two  intersecting  spheres  meets  the  surfaces 
of  the  spheres  in  the  poles  of  their  common  circle. 

11.  Find  the  locus  of  the  centers  of  all  spheres  tangent  to  a  given 
plane  at  a  given  point. 

12.  Three  spheres  each  of  radius  R  are  placed  on  a  horizontal  plane 
so  that  each  is  tangent  to  the  other  two.  A  fourth  sphere  of  the  same 
radius  is  placed  on  top  of  them  so  that  it  touches  each.  Find  the  dis- 
tance from  the  highest  point  of  the  top  sphere  to  the  plane. 


444  SOLID   GEOMETRY 

13.  If  a  spherical  triangle  is  isosceles,  its  polar  triangle  is  also 
isosceles. 

14.  Prove  that  the  volume  of  a  hollow  spherical  shell  whose  outer 

and  inner  radii  are  R  and  r,  respectively,  is  ^^i^  -  r){R'^ -[-  Rr  +  r^) 

o 

15.  If  a  surveyor  desires  that  the  sum  of  the  angles  in  any  spherical 
triangle  which  he  uses  on  the  earth's  surface  shall  be  within  one  minute 
of  180°,  what  is  the  largest  area  that  the  triangle  may  inclose? 

Suggestion.  —  See  §  489. 

16.  The  portion  of  a  sphere  bounded  by  the  planes  of  two  great 
semicircles  and  the  lune  which  they  form  on  the  surface  is  called  a 
spherical  wedge.  The  angle  between  the  planes  of  a  spherical  wedge  is 
15°,  and  the  diameter  of  the  sphere  is  12  in.  Find  the  volume  of  the 
wedge. 

17.  How  many  straight  lines  can  be  tangent  to  a  sphere  from  a  point 
outside  of  the  sphere  ?  Compare  their  lengths  between  the  given  point 
and  the  points  of  contact. 

18.  The  points  of  contact  of  all  lines  tangent  to  a  sphere  from  an 
external  point  lie  in  a  circle. 

19.  The  lines  in  Exercise  18  are  the  elements  of  a  cone. 

20.  If  two  spheres  are  tangent  to  the  same  plane  at  the  same  point, 
the  straight  line  joining  their  centers  passes  through  the  point  of 
contact.  - 

21.  What  is  the  locus  of  the  centers  of  all  spheres  tangent  to  a  given 
plane  at  a  given  point  ? 

22.  What  is  the  locus  of  the  centers  of  all  spheres  tangent  to  the 
faces  of  a  diedral  angle  ? 

23.  Find  the  edge  of  a  cube  inscribed  in  a  sphere  with  radius  10  in. 
Find  its  area.     Find  its  volume. 

24.  The  volume  of  a  polyedron  circumscribed  about  a  sphere  is  equal 
to  the  product  of  the  area  of  its  surface  and  one  third  of  the  radius  of 
the  sphere. 

25.  The  sum  of  the  arcs  of  great  circles  drawn  from  any  point  within 
a  spherical  triangle  to  the  extremities  of  a  side  is  less  than  the  sum  of 
the  other  two  sides  of  the  triangle. 

26.  If  from  a  point  within  a  spherical  triangle  arcs  of  great  circles  are 
drawn  to  the  three  vertices,  their  sum  is  less  than  the  perimeter. 


refere;nces  to  plane  geometry 

The  following  axioms,  theorems,  etc. ,  which  are  given  in  the  Plane  Geom- 
etry are  referred  to  in  the  proofs  in  the  Solid  Geometry.  They  are  placed 
here  for  the  convenience  of  the  student  in  looking  up  the  references. 

AXIOMS 

I.  Things  which  are  equal  to  the  same  thing,  or  to  equal  things,  are  equal 
to  each  other. 

II.  If  equals  are  added  to  equals,  the  sums  are  equal. 

III.  If  equals  are  subtracted  from  equals,  the  remainders  are  equal. 

IV.  If  equals  are  multiplied  by  equals,  the  products  are  equal. 

V.  If  equals  are  divided  by  equals,  the  quotients  are  equal. 
"VI.   Like  powers,  or  like  positive  roots,  of  equals  are  equal. 

VII.  If  equals  are  added  to  or  subtracted  from  unequals,  or  if  unequals 
are  multiplied  or  divided  by  the  same  positive  number,  the  results  are  unequal 
in  the  same  order. 

VIII.  If  unequals  are  subtracted  from  equals,  the  remainders  are  unequal 
in  the  reverse  order. 

IX.  If  unequals  are  added  to  unequals  in  the  same  order,  the  sums  are 
unequal  in  that  order. 

X.  The  whole  of  a  thing  is  equal  to  the  sum  of  all  of  its  parts,  and  is 
greater  than  any  one  of  its  parts. 

XL  If  the  first  of  three  things  is  greater  than  the  second,  and  the  second 
greater  than  the  third,  then  the  first  is  greater  than  the  third. 

XII.    A  quantity  may  be  substituted  for  its  equal  in  any  expression. 

DEFINITIONS,   THEOREMS,    ETC. 

§  6.  A  straight  line  is  represented  by  placing  a  ruler,  or  straightedge, 
upon  a  plane  surface  and  marking  along  the  edge  of  the  ruler. 

§  13.  Two  angles  are  equal  if,  and  only  if,  they  may  be  made  to  coincide 
throughout.  If  an  angle  is  placed  upon  an  equal  angle  so  that  the  vertices 
and  a  pair  of  sides  coincide,  the  other  sides  must  coincide. 

§  17.   A  right  angle  is  one  half  of  a  straight  angle. 

§  20.       I.   Any  two  straight  angles  are  equal. 
11.   Any  two  right  angles  are  equal. 

III.  The  complements  of  equal  angles  are  equal. 

IV.  The  supplements  of  equal  angles  are  equal. 
V.   Vertical  angles  are  equal. 

445 


446  REFERENCES  TO  PLANE   GEOMETRY 

§  26.  If  two  parallel  lines  are  cut  by  a  transversal,  the  corresponding  angles 
are  equal. 

§  34.  If  a  line  is  perpendicular  to  one  of  two  parallel  lines,  it  is  perpendic- 
ular to  the  other  also. 

§  41.   Two  straight  lines  perpendicular  to  the  same  straight  line  are  parallel. 

§  46.  Only  one  straight  line  can  be  drawn  through  a  given  point  parallel  to 
a  given  straight  line. 

§  48.  The  sum  of  the  angles  of  any  triangle  is  equal  to  a  straight  angle. 

§  53.  At  a  given  point  on  a  given  line  only  one  perpendicular  can  be 
drawn  to  the  line. 

§  54.  Only  one  perpendicular  to  a  given  line  can  be  drawn  through  a  given 
point  not  on  the  line. 

§  55.  Two  straight  lines  perpendicular  respectively  to  two  intersecting 
straight  lines  must  meet. 

§  60.  If  one  of  two  figures  may  be  placed  upon  the  other  so  that  they 
coincide  throughout,  the  figures  are  called  congruent. 

§  63.  If  two  sides  and  the  included  angle  of  one  triangle  are  equal  respec- 
tively to  two  sides  and  the  included  angle  of  another,  the  triangles  are  con- 
gruent. 

§  64.  Two  right  triangles  which  have  the  legs  of  one  equal  respectively  to 
the  legs  of  the  other  are  congruent. 

§  67.  Two  right  triangles  are  congruent  if  a  leg  and  an  acute  angle  of  one 
are  equal  respectively  to  a  leg  and  an  acute  angle  of  the  other. 

§  68.  Two  right  triangles  are  congruent  if  the  hypotenuse  and  an  acute 
angle  of  one  are  equal  respectively  to  the  hypotenuse  and  an  acute  angle  of 
the  other. 

§  76.  If  the  three  sides  of  one  triangle  are  equal  respectively  to  the  three 
sides  of  another,  the  triangles  are  congruent. 

§  82.   The  opposite  sides  of  a  parallelogram  are  equal. 

§  83.   A  diagonal  divides  a  parallelogram  into  two  congruent  triangles. 

§  85.   Two  parallel  lines  are  everywhere  equidistant. 

§  90.  If  two  opposite  sides  of  a  quadrilateral  are  equal  and  parallel,  the 
figure  is  a  parallelogram. 

§  100.  If  a  line-segment  is  parallel  to  the  bases  of  a  trapezoid  and  bisects 
one  of  the  non-parallel  sides,  then  it  bisects  the  other  also  and  is  equal  to  one 
half  of  the  sum  of  the  bases. 

§  101.  The  sum  of  the  interior  angles  of  a  convex  polygon  of  n  sides  is 
(n  —  2)  straight  angles. 

§  105.  The  locus  6f  points  equidistant  from  the  ends  of  a  line-segment  is 
the  perpendicular  bisector  of  it. 

§  107.  Two  points  each  equidistant  from  the  ends  of  a  line-segment  de- 
termine the  perpendicular  bisector  of  it. 


REFERENCES  TO  PLANE   GEOMETRY  447 

§  114.  The  medians  of  any  triangle  are  concurrent  at  a  point  of  trisection 
of  each. 

§  118.  (1)  If  four  numbers  are  in  proportion,  the  product  of  the  extremes 
equals  the  product  of  the  means. 

(2)  If  the  product  of  two  numbers  equals  the  product  of  two  other  num- 
bers, the  four  numbers  are  in  proportion,  one  pair  of  factors  being  extremes 
and  the  other  pair  means. 

(3)  If  four  numbers  are  in  proportion,  they  are  in  proportion  by  inversion. 

(4)  In  any  proportion,  the  means  may  be  interchanged,  or  the  extremes 
interchanged,  without  destroying  the  proportion. 

(6)  The  terms  of  any  proportion  are  in  proportion  by  addition. 

(6)  The  terms  of  any  proportion  are  in  proportion  by  subtraction. 

(7)  Like  powers  or  like  roots  of  the  terms  of  a  proportion  are  in  proportion. 

(8)  If  two  or  more  ratios  are  equal,  the  sum  of  the  antecedents  is  to  the 
sum  of  the  consequents  as  any  antecedent  is  to  its  consequent. 

§  123.  If  DE  is  parallel  to  side  AB  of  triangle  ABC  and  meets  AC  a.t  D 
and  BC  at  E^  then 

^  =  ^and^=^. 
AD      BE  DC     EC 

§  127.  Two  polygons  which  have  the  angles  of  one  equal  respectively  to 
the  angles  of  the  other,  taken  in  order,  are  called  mutually  equiangular. 

Similar  polygons  are  those  which  (1)  are  mutually  equiangular  and 
(2)  have  their  corresponding  sides  proportional. 

§  128.   If  two  triangles  are  mutually  equiangular,  they  are  similar. 

§  129.  If  two  triangles  have  their  corresponding  sides  proportional,  the 
triangles  are  similar. 

§  130.  If  two  triangles  have  an  angle  of  one  equal  to  an  angle  of  the  other, 
and  the  including  sides  proportional,  they  are  similar. 

§  146.   The  sum  of  any  two  sides  of  a  triangle  is  greater  than  the  third  side. 

§  148.  If  two  triangles  have  two  sides  of  one  equal  respectively  to  two  sides 
of  the  other,  but  the  third  side  of  the  first  greater  than  the  third  side  of  the 
second,  the  angles  opposite  these  sides  are  unequal,  the  greater  angle  being 
opposite  the  greater  side. 

§  151.    (1)  A  diameter  of  a  circle  equals  two  times  a  radius. 

(2)  Radii  of  the  same  circle  or  equal  circles  are  equal. 

(3)  If  the  radii  of  two  circles  are  equal,  the  circles  are  equal. 

(4)  In  the  same  or  equal  circles,  equal  central  angles  intercept  equal  arcs. 

(5)  In  the  same  or  equal  circles,  equal  arcs  subtend  equal  central  angles. 

(6)  In  the  same  or  equal  circles,  equal  arcs  are  subtended  by  equal  chords. 

(7)  In  the  same  or  equal  circles,  equal  chords  subtend  equal  arcs. 

(8)  A  point  is  at  a  distance  from  the  center  of  a  circle  equal  to,  greater 


448  REFERENCES  TO  PLANE   GEOMETRY 

tlian,  or  less  than  the  radius,  according  as  it  is  on,  outside  of,  or  within  the 
circle ;  and  conversely. 

§  156.  A  line  through  the  center  of  a  circle  perpendicular  to  a  chord 
bisects  the  chord  and  the  subtended  arcs. 

§  164.  A  tangent  to  a  circle  is  perpendicular  to  the  radius  drawn  to  the 
point  of  contact. 

§  173.  A  central  angle  has  the  same  numerical  measure  as  its  intercepted 
arc. 

§  185.  If,  in  geometric  constructions,  the  instruments  used  are  limited  to 
the  ungraduated  straightedge  and  compasses  alone,  there  are  some  construc- 
tions which  cannot  be  made.  There  are  three  such  impossible  constructions 
which  have  interested  mathematicians  since  the  time  of  the  ancient  Greeks  : 

(1)  To  trisect  (cut  into  three  equal  parts)  any  given  angle. 

(2)  To  construct  a  square  which  shall  have  the  same  area  as  a  given  circle. 

(3)  To  construct  a  cube  which  shall  have  twice  the  volume  of  a  given  cube. 
§  196.   In  any  right  triangle,  the  square  of  the  hypotenuse  is  equal  to  the 

sum  of  the  squares  of  the  legs. 

§  200.  If  two  chords  intersect,  the  product  of  the  segments  of  one  is  equal 
to  the  product  of  the  segments  of  the  other. 

§  215.  Parallelograms  having  equal  altitudes  and  equal  bases  are  equal. 

§  228.  Two  triangles  having  an  angle  of  one  equal  to  an  angle  of  the  other 
are  to  each  other  as  the  products  of  the  sides  including  those  angles. 

§  274.  (1)  If  the  number  of  sides  of  a  regular  inscribed  polygon  is  indefi- 
nitely increased,  the  apothem  approaches  the  radius  of  the  circle  as  a  limit. 

(2)  If  the  number  of  sides  of  a  regular  inscribed  polygon  is  indefinitely 
increased,  the  perimeter  approaches  the  circumference  of  the  circle  as  a  limit. 

(3)  If  the  number  of  sides  of  a  regular  circumscribed  polygon  is  indefi- 
nitely increased,  the  perimeter  of  the  polygon  approaches  the  circumference 
of  the  circle  as  a  limit. 

(4)  If  the  number  of  sides  of  a  regular  circumscribed  polygon  is  indefi- 
nitely increased,  the  area  of  the  polygon  approaches  the  area  of  the  circle  as 
a  limit. 

§  275.  (1)  If  two  variables  are  equal  and  each  approaches  a  limit,  the 
limits  are  equal. 

(2)  If  the  limit  of  a  variable  x  is  a,  then  the  limit  of  Tex  is  Tca^  where  Tc  is 
any  constant. 

(3)  If  the  limit  of  a  variable  x  is  a,  then  the  limit  of  -  is  -,  where  k  is 

Tc       k 
any  constant. 

§  281.   The  area  of  a  circle  equals  vB^. 

§  284.   The  area  of  a  sector  of  a  circle  is  ecjual  to  one  half  of  the  product 

of  its  radius  and  its  arc. 


SOLID   GEOMETRY 
INDEX   OF   DEFINITIONS 


Altitude  of  a  cone,  §  398. 

of  a  cylinder,  §  371. 

of  a  frustum  of  a  cone,  §  405. 

of  a  frustum  of  a  pyramid,  §  385. 

of  a  prism,  §  349. 

of  a  prismatoid,  §  429. 

of  a  pyramid,  §  383. 

of  a  spherical  segment,  §  457. 

of  a  zone,  §  450. 
Angle,  diedral,  §  325. 

of  a  spherical  polygon,  §  464. 

polyedral,  §  341. 

right  spherical,  §  459. 

spherical,  §  459. 

triedral,  §  341. 
Arcs,  perpendicular,  §  460. 
Axis  of  a  circle,  §  436. 

Base  of  a  cone,  §  398. 

of  a  pyramid,  §  383. 

of  a  spherical  sector,  §  455. 
Bases  of  a  cylinder,  §  371. 

of  a  frustum  of  a  cone,  §  405. 

of  a  prism,  §  349. 

of  a  prismatoid,  §  429. 

of  a  spherical  segment,  §  457. 

of  a  truncated  pyramid,  §  384. 

of  a  zone,  §  450. 

Center  of  a  sphere,  §  432. 
Cone,  §  398. 

circular,  §  398. 

of  revolution,  §  398. 

right  circular,  §  398. 
Cones,  similar,  §  398, 
Cube,  §  352. 
Cylinder,  §  371. 

circular,  §  371. 

circumscribed,  §  374, 

inscribed,  §  374. 

oblique,  §  371. 

of  revolution,  §  378. 

right,  §  371. 

right  circular,  §  371. 


Cylinders,  similar,  §  378. 

Degree,  spherical,  §  486. 
Determining  a  plane,  §  295. 
Diagonal  of  a  polyedron,  §  348. 

of  a  spherical  polygon,  §  464. 
Diameter  of  a  sphere,  §  432. 
Diedral  angle,  §  325. 

plane  angle  of,  §  326. 

right,  acute,  obtuse,  §  329. 
Diedral  angles,  adjacent,  §  329. 

complementary,  §  329. 

supplementary,  §  329. 

vertical,  §  329. 
Dimensions  of  a  parallelopiped,  §  363. 
Distance  between  parallel  planes,  §  322. 

from  a  point  to  a  plane,  §  309. 

on  the  surface  of  a  sphere,  §  437. 

polar,  §  439. 
Dodecaedron,  §  348. 

Edges  of  a  polyedral  angle,  §  341. 

of  a  polyedron,  §  348. 
Element  of  a  conical  surface,  §  398. 

of  a  cylindrical  surface,  §  371. 
Ellipse,  §  404. 
Excess,  spherical,  of  spherical  polygon,  §  49Q 

spherical,  of  spherical  triangle,  §  476. 

Faces  of  a  polyedral  angle,  §  341. 

of  a  polyedron,  §  348. 
Frustum  of  a  pyramid,  §  385. 

circumscribed,  §  406. 

inscribed,  §  406. 

of  a  cone,  §  405. 

of  a  cone,  inscribed,  §  406. 

Geometry,  solid,  §  288. 
Great  circle  of  a  sphere,  §  435. 

Hexaedron,  §  348. 
Hyperbola,  §  404. 

Icosaedron,  §  348. 

Inclination  of  line  to  plane,  §  340. 


449 


450 


INDEX  OF  DEFINITIONS 


Lateral  area  of  a  prism,  §  349, 

of  a  pyramid,  §  383. 
Lateral  edges  of  a  prism,  §  349. 

of  a  pyramid,  §  383. 
Lateral  faces  of  a  prism,  §  349. 

of  a  pyramid,  §  383. 
Lateral  surface  of  a  cone,  §  398. 

of  a  cylinder,  §  371. 
Line  oblique  to  a  plane,  §  302. 

perpendicular  to  a  plane,  §  302. 
Lune,  §  487. 

Mid-section  of  a  prismatoid,  §  429. 

Nappes  of  a  conical  surface,  §  398. 

Octaedron,  §  348. 

Parabola,  §  404. 

Parallel  line  and  plane,  §  312. 

planes,  §  312. 
Parallelopiped,  §  352. 

rectangular,  §  352. 

right,  §  352. 
Perpendicular  planes,  §  330. 
Plane  angle  of  diedral  angle,  §  326. 
Plane  perpendicular  to  a  line,  §  302. 

tangent  to  a  cylinder,  §  374. 
Planes,  parallel,  §  312. 

perpendicular,  §  330. 
Poles  of  a  circle,  §  436. 
Polyedral  angles,  §  341. 

convex,  §  341. 

equal  and  symmetrical,  §  344. 
Polyedrons,  §  348. 

regular,  §  418. 

similar,  §  421. 
Polygon,  spherical,  §  464. 

convex,  §  464. 
Projection  of  line  upon  plane,  §  338. 

of  point  upon  plane,  §  338. 
Prism,  §  349. 

circumscribed,  §  374. 

inscribed,  §  374. 

oblique,  §  349. 

quadrangular,  §  349. 

right,  §  349. 

triangular,  §  349. 
Prismatoid,  §  429. 
Pyramid,  §  383. 

circumscribed,  §  406. 

hexagonal,  §  383. 

inscribed,  §  406. 

quadrangular,  §  383. 


regular,  §  383. 
triangular,  §  383. 

Radius  of  a  sphere,  §  432. 
Right  section  of  a  cylinder,  §  371. 
of  a  prism,  §  349. 

Section,  conic,  §  404. 

of  a  cylinder,  §  371. 

of  a  polyedron,  §  348. 
Sector,  spherical,  §  455. 
Segment,  spherical,  §  457. 

of  one  base,  §  457. 
Slant  height  of  a  cone,  §  400. 

of  a  frustum  of  a  cone,  §  405. 

of  a  frustum  of  a  pyramid,  §  387< 

of  a  pyramid,  §  387. 
Small  circle  of  a  sphere,  §  435. 
Solids,  congruent,  §  357. 

equal,  §  357. 

geometric,  §  347. 
Sphere,  §  432. 

circumscribed,  §  445. 

inscribed,  §  445. 
Surface,  conical,  §  398. 

cylindrical,  §  371. 

spherical,  §  432. 

Tangent  to  a  sphere,  §  442. 
Tetraedron,  §  348. 
Triangle,  spherical,  §  464. 

equilateral,  §  464. 

isosceles,  §  464. 

right,  §  464. 
Triangles,  birectangular,  §  476. 

mutually  equiangular,  §  472. 

mutually  equilateral,  §  472. 

polar,  §  468. 

symmetrical,  §  477. 

trirectangular,  §  476. 
Triedral  angle,  §  341. 
Truncated  prism,  §  351. 

pyramid,  §  384. 

Vertex  of  a  conical  surface,  §  398. 

of  a  polyedral  angle,  §  341. 

of  a  pyramid,  §  383. 

of  a  spherical  angle,  §  459. 
Vertices  of  a  polyedron,  §  348. 

of  a  spherical  polygon,  §  464. 
Volume  of  a  solid,  §  356. 

Zone,  §  450. 

of  one  base,  §  450. 


14  DAY  USE 

RETURN  TO  DESK  FROM  WHICH  BORROWED 

LOAN  DEPT. 

This  book  is  due  on  the  last  date  stamped  below,  or 

on  the  date  to  which  renewed. 

Renewed  books  are  subject  to  immediate  recall. 


RECD  LD 

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RECCIR.  JUL     1  '80 

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(A1724sl0)476B 

General  Library 

University  of  California 

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LD21-100m-7,'40  (6936s) 

THE  UNIVERSITY  OF  CAUFORNIA  LIBRARY 


